what is the equation of a quadratic function p with rational coefficients that has a zero of 3-i

Answer:

(a) Probability that a ball bearing is between the target and the actual mean is 0.2734.

(b) Probability that a ball bearing is between the lower specification limit and the target is 0.226.

(c) Probability that a ball bearing is above the upper specification limit is 0.0401.

(d) Probability that a ball bearing is below the lower specification limit is 0.0006.

Step-by-step explanation:

We are given that an industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specification limits under which the ball bearings can operate are 0.74 inch and 0.76 inch, respectively.

Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed, with a mean of 0.753 inch and a standard deviation of 0.004 inch.

Let X = diameter of the ball bearings

SO, X ~ Normal([tex]\mu=0.753,\sigma^{2} =0.004^{2}[/tex])

The z-score probability distribution for normal distribution is given by;

                                Z  =  [tex]\frac{X-\mu}{\sigma} } }[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = population mean = 0.753 inch

           [tex]\sigma[/tex] = standard deviation = 0.004 inch

(a) Probability that a ball bearing is between the target and the actual mean is given by = P(0.75 < X < 0.753) = P(X < 0.753 inch) - P(X [tex]\leq[/tex] 0.75 inch)

      P(X < 0.753) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] < [tex]\frac{0.753-0.753}{0.004} } }[/tex] ) = P(Z < 0) = 0.50

      P(X [tex]\leq[/tex] 0.75) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] [tex]\leq[/tex] [tex]\frac{0.75-0.753}{0.004} } }[/tex] ) = P(Z [tex]\leq[/tex] -0.75) = 1 - P(Z < 0.75)

                                                             = 1 - 0.7734 = 0.2266

The above probability is calculated by looking at the value of x = 0 and x = 0.75 in the z table which has an area of 0.50 and 0.7734 respectively.

Therefore, P(0.75 inch < X < 0.753 inch) = 0.50 - 0.2266 = 0.2734.

(b) Probability that a ball bearing is between the  lower specification limit and the target is given by = P(0.74 < X < 0.75) = P(X < 0.75 inch) - P(X [tex]\leq[/tex] 0.74 inch)

      P(X < 0.75) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] < [tex]\frac{0.75-0.753}{0.004} } }[/tex] ) = P(Z < -0.75) = 1 - P(Z [tex]\leq[/tex] 0.75)

                                                            = 1 - 0.7734 = 0.2266

      P(X [tex]\leq[/tex] 0.74) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] [tex]\leq[/tex] [tex]\frac{0.74-0.753}{0.004} } }[/tex] ) = P(Z [tex]\leq[/tex] -3.25) = 1 - P(Z < 3.25)

                                                             = 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 0.75 and x = 3.25 in the z table which has an area of 0.7734 and 0.9994 respectively.

Therefore, P(0.74 inch < X < 0.75 inch) = 0.2266 - 0.0006 = 0.226.

(c) Probability that a ball bearing is above the upper specification limit is given by = P(X > 0.76 inch)

      P(X > 0.76) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] > [tex]\frac{0.76-0.753}{0.004} } }[/tex] ) = P(Z > -1.75) = 1 - P(Z [tex]\leq[/tex] 1.75)

                                                            = 1 - 0.95994 = 0.0401

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.95994.

(d) Probability that a ball bearing is below the lower specification limit is given by = P(X < 0.74 inch)

      P(X < 0.74) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] < [tex]\frac{0.74-0.753}{0.004} } }[/tex] ) = P(Z < -3.25) = 1 - P(Z [tex]\leq[/tex] 3.25)

                                                            = 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 3.25 in the z table which has an area of 0.9994.

Using the normal distribution, it is found that there is a

a) 0.2734 = 27.34% probability that a ball bearing is between the target and the actual mean.

b) 0.226 = 22.6% probability that a ball bearing is between the lower specification limit and the target.

c) 0.0401 = 4.01% probability that a ball bearing is above the upper specification limit.

d) 0.0006 = 0.06% probability that a ball bearing is below the lower specification limit.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

In this problem:

Item a:

This probability is the p-value of Z when X = 0.753 subtracted by the p-value of Z when X = 0.75.

X = 0.753:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.753 - 0.753}{0.004}[/tex]

[tex]Z = 0[/tex]

[tex]Z = 0[/tex] has a p-value of 0.5

X = 0.75:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.75 - 0.753}{0.004}[/tex]

[tex]Z = -0.75[/tex]

[tex]Z = -0.75[/tex] has a p-value of 0.2266

0.5 - 0.2266 = 0.2734

0.2734 = 27.34% probability that a ball bearing is between the target and the actual mean.

Item b:

This probability is the p-value of Z when X = 0.75 subtracted by the p-value of Z when X = 0.74, hence:

X = 0.75:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.75 - 0.753}{0.004}[/tex]

[tex]Z = -0.75[/tex]

[tex]Z = -0.75[/tex] has a p-value of 0.2266

X = 0.74:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.74 - 0.753}{0.004}[/tex]

[tex]Z = -3.25[/tex]

[tex]Z = -3.25[/tex] has a p-value of 0.0006.

0.2266 - 0.0006 = 0.226

0.226 = 22.6% probability that a ball bearing is between the lower specification limit and the target.

Item c:

This probability is 1 subtracted by the p-value of Z when X = 0.76, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.76 - 0.753}{0.004}[/tex]

[tex]Z = 1.75[/tex]

[tex]Z = 1.75[/tex] has a p-value of 0.9599.

1 - 0.9599 = 0.0401

0.0401 = 4.01% probability that a ball bearing is above the upper specification limit.

Item d:

This probability is the p-value of Z when X = 0.74, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0.74 - 0.753}{0.004}[/tex]

[tex]Z = -3.25[/tex]

[tex]Z = -3.25[/tex] has a p-value of 0.0006.

0.0006 = 0.06% probability that a ball bearing is below the lower specification limit.

A similar problem is given at https://brainly.com/question/24663213

Answer:

A. When its sides are congruent

Step-by-step explanation:

By definition a square is a four sided polygon (shape) which has four right angles and four sides of equal length.

So, when a rectangle has all congruent sides, its a square because its a rectangle so it has four right angles, and congruent sides are sides of the same length.

B is incorrect because a rectangle can have four right angles but have differing side lengths

C is incorrect because again a rectangle can have parallel sides but have differing side lengths

D is incorrect because it can have convex angles but again have differing side lengths

When sides of rectangle are congruent, then it is a square. Therefore, the correct answer is option A.

A rectangle is a closed two-dimensional figure with four sides. The opposite sides of a rectangle are equal and parallel to each other and all the angles of a rectangle are equal to 90°.

A square is a quadrilateral with four equal sides. There are many objects around us that are in the shape of a square. Each square shape is identified by its equal sides and its interior angles that are equal to 90°.

If all the sides of rectangle are congruent, then it is square.

Therefore, the correct answer is option A.

Learn more about the square here:

brainly.com/question/28776767.

#SPJ4

Your question is not well presented (Refer below for correct question)

The distance it takes a truck to stop can be modeled by the function;

d(u) = 2.15u²/64.4f

Where

d = stopping distance in feet

u = initial velocity in miles per hour

f = a constant related to friction

When the truck's initial velocity on dry pavement is 40 mph, its stopping distance is 138 ft.

Determine the value of f.

Answer:

The friction constant is approximately 0.3871

Step-by-step explanation:

Given

Stopping distance, d(u) = 138ft

Initial Velocity, u = 40mph

Required

Friction constant, f.

To get the value of f, we need to simply substitute 40 for u and 138 for d(u) in the above expression.

In other words;

d(u) = 2.15u²/64.4f becomes

138 = 2.15 * 40²/64.4f

138 = 2.15 * 1600/64.4f

138 = 3440/64.4f

Multiply both sides by 64.4f

138 * 64.4f = 64.4f * 3440/64.4f

8887.2f = 3440

Divide both sides by 8887.2

8887.2f/8887.2 = 3440/8887.2

f = 3440/8887.2

f = 0.387073544

f = 0.3871 (Approximated)

Hence, the friction constant is approximately 0.3871

Answer:

first one is  .39
second answer is C   the last  one

Step-by-step explanation:

Answer:

See explanation

Step-by-step explanation:

Q1)a

- Denote a random variable ( X ) as the life time of a brand of bulb produced.

- The given mean ( μ ) = 210 hrs and standard deviation ( σ ) = 56 hrs. The distribution is symbolized as follows:

                           X ~ Norm ( 210 , 56^2 )

i) The bulb picked to have a life time of at least 300 hours.

- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:

          P ( X ≥ x ) = P ( Z ≥ ( x - μ ) / σ )

          P ( X ≥ 300 ) = P ( Z ≥ ( 300 - 210 ) / 56 )

          P ( X ≥ 300 ) = P ( Z ≥ 1.607 )  

- Use the standard normal look-up table for limiting value of Z-score:

         P ( X ≥ 300 ) = P ( Z ≥ 1.607 ) = 0.054  .. Answer

ii) The bulb picked to have a life time of at most 100 hours.

- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:

          P ( X ≤ x ) = P ( Z ≤ ( x - μ ) / σ )

          P ( X ≤ 100 ) = P ( Z ≤ ( 100 - 210 ) / 56 )

          P ( X ≤ 100 ) = P ( Z ≤ -1.9643 )  

- Use the standard normal look-up table for limiting value of Z-score:

         P ( X ≤ 100 ) = P ( Z ≤ -1.9643 ) = 0.0247  .. Answer

iii) The bulb picked to have a life time of between 150 and 250 hours.

- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:

          P ( x1 ≤ X ≤ x2 ) = P ( ( x1 - μ ) / σ ≤ Z ≤ ( x2 - μ ) / σ )

          P ( 150 ≤ X ≤ 250 ) = P ( ( 150 - 210 ) / 56 ≤ Z ≤ ( 250 - 210 ) / 56 )

          P ( 150 ≤ X ≤ 250 ) = P ( -1.0714 ≤ Z ≤ 0.71428 )  

- Use the standard normal look-up table for limiting value of Z-score:

         P ( 150 ≤ X ≤ 250 ) = P ( -1.0714 ≤ Z ≤ 0.71428 ) = 0.6205  .. Answer

Q1)b

- Denote event (A) : Kofi solves the problem correctly. Then the probability of him answering successfully is:

                    p ( A ) = 0.25

- Denote event (B) : Menesh solves the problem correctly. Then the probability of him answering successfully is:

                    p ( B ) = 0.4

- The probability that neither of them answer the question correctly is defined by a combination of both events ( A & B ). The two events are independent.  

- So for independent events the required probability can be stated as:

              p ( A' & B' ) = p ( A' ) * p ( B' )

              p ( A' & B' ) = [ 1 - p ( A ) ] * [ 1 - p ( B ) ]

              p ( A' & B' ) = [ 1 - 0.25 ] * [ 1 - 0.4 ]

              p ( A' & B' ) = 0.45 ... Answer

Q2)a

- A discrete random variable X: defines the probability of getting each number on a biased die.

- From the law of total occurrences. The sum of probability of all possible outcomes is always equal to 1.

             ∑ p ( X = xi ) = 1

             p ( X = 1 ) + p ( X = 2 ) + p ( X = 3 ) + p ( X = 4 ) + p ( X = 5 ) + p ( X = 6 )

             1/6 + 1/6 + 1/5 + k + 1/5 + 1/6 = 1

             k = 0.1  ... Answer

- The expected value E ( X ) or mean value for the discrete distribution is determined from the following formula:

             E ( X ) = ∑ p ( X = xi ) . xi

             E ( X ) = (1/6)*1 + (1/6)*2 + (1/5)*3 + (0.1)*4 + (1/5)*5 + (1/6)*6

             E ( X ) = 3.5 .. Answer

- The expected-square value E ( X^2 ) or squared-mean value for the discrete distribution is determined from the following formula:

             E ( X^2 ) = ∑ p ( X = xi ) . xi^2

             E ( X^2 ) = (1/6)*1 + (1/6)*4 + (1/5)*9 + (0.1)*16 + (1/5)*25 + (1/6)*36

             E ( X^2 ) = 15.233 .. Answer

- The variance of the discrete random distribution for the variable X can be determined from:

            Var ( X ) = E ( X^2 ) - [ E ( X ) ] ^2

            Var ( X ) = 15.2333 - [ 3.5 ] ^2

            Var ( X ) = 2.9833 ... Answer

- The cumulative probability of getting any number between 1 and 5 can be determined from the sum:

           P ( 1 < X < 5 ) = P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 )

           P ( 1 < X < 5 ) = 1/6 + 1/5 + 0.1

           P ( 1 < X < 5 ) = 0.467  ... Answer

Q2)b

- Two independent events are defined by their probabilities as follows:

           p ( A ) = 0.3  and p ( B ) = 0.5

- The occurrences of either event does not change alter or affect the occurrences of the other event; hence, independent.

- For the two events to occur simultaneously at the same time:

          p ( A & B ) = p ( A )* p ( B )

          p ( A & B ) = 0.3*0.5

          p ( A & B ) = 0.15  ... Answer

- For either of the events to occur but not both. From the comparatively law of two independent events A and B we have:

        p ( A U B ) = p ( A ) + p ( B ) - 2*p ( A & B )

        p ( A U B ) = 0.3 + 0.5 - 2*0.15

        p ( A U B ) = 0.5 ... Answer

- Two mutually exclusive events can-not occur simultaneously; hence, the two events are not mutually exclusive because:

       p ( A & B ) = 0.15 ≠ 0

Q2)c

- The letters of the word given are to be arranged in number of different ways as follows:

                                  STATISTICS

- Number of each letters:

        S : 3

        T : 3

        A: 1

        I: 2

        C: 1

- 10 letters can be arranged in 10! ways.

- However, the letters ( S and T and I ) are repeated. So the number of permutations must be discounted by the number of each letter is repeated as follows:

                        [tex]\frac{10!}{3!3!2!} = \frac{3628800}{72} = 50,400[/tex]

- So the total number of ways the word " STATISTICS " can be re-arranged is 50,400 without repetitions.

Answer:

  1a) 210

  1b) 0.18

  2) 17.7 will fit (> 17)

  3) 515 cm^2

Step-by-step explanation:

1a) The sample of 25 is exactly 1/30 of the population of 750, so the number of orange cakes in the population will be 30 times the number in the sample.

The total number of orange cakes made on Monday is 30·7 = 210.

__

1b) The lowest number that will round to 5 is 4.5. The lowest probability that a cake is an orange cake is ...

  4.5/25 = 18/100 = 0.18

The lower bound on the probability that the cake is orange is 0.18.

____

2) The working shown is partly correct.

The possible error in the box dimension is 1/2 cm, so the box may be as short as 28 -1/2 cm = 27.5 cm.

The possible error in the disc case dimension is 1/2 mm = 0.05 cm, so the case may be as thick as 1.5 +.05 = 1.55 cm.

The minimum number discs that will fit in the case is (27.5 cm)/(1.55 cm) = 17.7. Thus, 17 cases will definitely fit in the box.

____

3) The shaded area will be the largest when the outside rectangle is the largest and the inside rectangle is the smallest. Errors in dimensions can be as much as 1/2 cm, so the Outside rectangle can be 42.5 cm by 21.5 cm. The inside rectangle can be as small as 27.5 cm by 14.5 cm.

Then the shaded area can be as large as ...

  (42.5)(21.5) -(27.5)(14.5) = 515 . . . . cm^2

Answer:

h = 6

Step-by-step explanation:

We need to use Pythagorean theorem.

h² + 8² = 10²

h² = 100 - 64 = 36

h = 6

Answer:

y>5, x+y underlined<12

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

The first one will be da answer

Step-by-step explanation:

happy to help you!

Answer:

1)   y = x + 3

2)   31 years old

Step-by-step explanation:

Jack is x years old

Todd is y years old

Todd is 3 years older than his brother Jack => y = x + 3

When Jack is 28 years old, Todd is y = 28 + 3 = 31 years old

Answer:

B

Step-by-step explanation:

Since you are given line RT and ND, then all you need to do by ASA is the make sure that the angles are the two endpoints are congruent. Since the problem already gave you R and N, then all thats left is to relate the other two endpoints, namely, T and D

Answer:60°

Step-by-step explanation:I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thank me....

Answer:

A.

60°

Hope its right.

Step-by-step explanation:

Answer:

5y - 4 ≥ 26

5y ≥ 26 + 4

5y ≥ 30

y ≥ 30/5

y ≥ 6

Answer:

4x^3 + 4x^2

Step-by-step explanation:

f(x) = 4x^2

g(x) = x + 1

=> (f.g)(x) = 4x^2*(x + 1) = 4x^3 + 4x^2

Hope this helps!

Answer:

2.75 inches

Step-by-step explanation:

City B experienced three times the amount of rainfall than  City A, therefore :

Since both cities received 11 inches of rainfall in total.

We have that:

x+3x=11

4x=11

Divide both sides by 4

x=2.75 Inches

Therefore, we City A received 2.75 inches of rain.

Answer:

0.26

Step-by-step explanation:

Answer: 1.1

Step-by-step explanation:

If Sam buys 14 water bottles, he will have spent 18.9, so when you subtract 20 (the limit) from 18.9 (how much money he'll spend on the water bottles) you get 1.1

Answer:

Y=-4/5x+3/5y

Step-by-step explanation: first get the gradient

3--1/-3-2

=4/-5

y-3/x+3=4/-5

-5y+15=4x+12

-5y=4x-3

y=4/-5x+3/5

Step-by-step explanation:

lateral area=area of the full triangle

surface area=area of the four triangles + area of the square

Answer:

It’s C

Step-by-step explanation:

But thanks for helping me get it wrong toxo360❤️

In the regression equation, Year 18 is the best prediction for when 2528 people will attend graduate school.

The polynomial equation whose highest degree is two is called a quadratic equation or sometimes just quadratics. It is expressed in the form of: ax² + bx + c = 0.

The number of people attending graduate school at a university may be modeled by the quadratic regression equation y = 10x2 - 40x+8, where x represents the year.

The regression equation, which year is the best prediction for when 2528 people will attend graduate school is;

[tex]\rm y = 10x^2 - 40x+8\\\\2528 = 10x^2 - 40x+8\\\\ 10x^2 - 40x+8\\\\ 10x^2 - 40x+8-2528=0 \\\\ 10x^2 - 40x-2520=0\\\\\text{Divide by 10 both sides}\\\\x^2-4x-252=0\\\\x^2-18x+14x-252=0\\\\x(x-18)+14(x-18)=0\\\\(x-18)(x+14)=0\\\\x-18=0, \ x=18\\\\x+14=0, \ x=-14[/tex]

Hence, the regression equation, Year 18 is the best prediction for when 2528 people will attend graduate school.

Learn more about quadratic equations here;

https://brainly.com/question/24298861

#SPJ2

Answer:

hope this is correct

Answer: after 6 mins, sam is 60 meters from the dock

Step-by-step explanation:

Answer:

after 6 mins, sam is 60 meters from the dock

Step-by-step explanation:

Answer:

-26

Step-by-step explanation:

y=mx+c

m= (y2 - y1 ) / (x2-x1)

m= (10-1) / (48 -36)

m = 0.75

10 = 0.75(48) + c

c = 10 - 36

c = -26

Answer:

$1946

Step-by-step explanation:

His average for the first four months

= $1,450.25

So he earned 4*$1,450.25

= $5801 for the first four months.

Then for the year his average is $1,780.75.

So he earned $1,780.75*12 for the year.

=$ 21369

So the amount he earned for the remaining 8 months was 21369-5801

= $15568

The average for the 8 Months= 15568/8

= $1946

Answer:

5x^2 -4

Step-by-step explanation:

f(x)= 4x^2+ 1

g(x) = x^2 - 5

(f+g)(x) = f(x) + g(x)

= 4x^2+ 1 + x^2 - 5

= 4x^2 + x^2 + 1- 5

= 5x^2 -4

Answer:

Step-by-step explanation:

The question is incomplete. The complete question is:

EU (European Union) countries report that 46% of their labor force is female. The United Nations wants to determine if the percentage of females in the U.S. labor force is the same. Based on a sample of 500 employment records, representatives from the United States Department of Labor found that the 95% confidence interval for the proportion of females in the U.S. labor force is 0.357 to 0.443. If the Department of Labor wishes to tighten its interval, they should:

A. increase the confidence level

B. decrease the sample size

C. increase the sample size

D. Both A and B

E. Both A and C

Solution:

Confidence interval for population proportion is written as

Sample proportion ± margin of error

Where sample proportion is the point estimate for the population proportion.

Margin of error = z × √pq/n

The z score for 95% confidence level is 1.96

p = 46/100 = 0.46

q = 1 - p = 1 - 0.46

q = 0.54

n = 500

Margin of error = 1.96√0.46 × 0.54/500 = 0.044

To tighten it's interval, the margin of error needs to be reduced.

If we increase the confidence level, say to 99%, z = 2.58

Then

Margin of error = 2.58√0.46 × 0.54/500 = 0.058

It increased

Also, If we increase the sample size, say to 700, then

Margin of error = 1.96√0.46 × 0.54/700 = 0.037

It has reduced

Therefore, the correct options is

C. increase the sample size

Answer:

72%

Step-by-step explanation:

As the T-shirts have 20% off and Juanita would get an extra 10% off the sale price, the additional discount would be over the price with the 20% off included. This means that the sale price is the 80% of the normal price and an additional 10% over that 80% would be:

80*10%= 8

80%-8%= 72%

This means that Juanita would have to pay the 72% percent of the regular price of the T-shirt as the total discount would be 28% off.

Answer:

Step-by-step explanation:

Total Investment = ​$100,000

The investment was split into three parts (say x, y and z)

Simple Interest = Principal X Rate/100 X Time

The first part of the investment earned​ 8% interest

Interest on the first part = 0.08x

The second part of the investment earned​ 6% interest

Interest on the second  part = 0.06y

The third part of the investment earned​ 9% interest

Interest on the third part = 0.09z

Total interest from the investments = $ 7380.

Therefore:

0.08x+0.06y+0.09z=7380

The interest from the first investment was 2 times the interest from the second.

0.08x=2 X 0.06y

0.08x=0.12y

Substituting we have:

0.12y+0.06y+0.09z=7380

0.18y+0.09z=7380

From 0.08x=0.12y

x=1.5y

Substituting x=1.5y into x+y+z=100000

1.5y+y+z=100000

2.5y+z=100000

z=100000-2.5y

Substituting z=100000-2.5y into 0.18y+0.09z=7380

0.18y+0.09(100000-2.5y)=7380

0.18y+9000-0.225y=7380

-0.045y=-1620

Divide both sides by -0.045

y=36000

Recall: x=1.5y

x=1.5 X 36000

x =54000

x+y+z=100000

54000+36000+z=100000

z=100000-(54000+36000)

z=10,000

Therefore:

Answer:

About 10 i think

Step-by-step explanation:

idk

Answer:

Jennifer because she would make less but spent more on supplies

Step-by-step explanation:

Answer: The answer is D! Jennifer, |-85| > |-60|

Step-by-step explanation: I did the quiz hope everyone does great ^^

Answer:

The base is 6cm

Step-by-step explanation:

Using [tex]\frac{1}{2}[/tex]bh=area, you just plug in 6 for h and 18 for area and solve. The formula should be 3b=18, which equals 6

The solution is Option D.

The length of the base of the triangle is B = 6 cm

A triangle is a plane figure or polygon with three sides and three angles.

A Triangle has three vertices and the sum of the interior angles add up to 180°

Let the Triangle be ΔABC , such that

∠A + ∠B + ∠C = 180°

The area of the triangle = ( 1/2 ) x Length x Base

For a right angle triangle

From the Pythagoras Theorem , The hypotenuse² = base² + height²

if a² + b² = c² , it is a right triangle

if a² + b² < c² , it is an obtuse triangle

if a² + b² > c² , it is an acute triangle

Given data ,

Let the area of the triangle be represented as A

Now , the value of A = 18 cm²

Let the base of the triangle be B

Let the height of the triangle be H = 6 cm

Now , area of the triangle = ( 1/2 ) x Length x Base

Substituting the values in the equation , we get

18 = ( 1/2 ) x B x 6

18 = 3B

Divide by 3 on both sides of the equation , we get

B = 6 cm

Hence , the base length of the triangle is B = 6 cm

To learn more about triangles click :

https://brainly.com/question/16739377

#SPJ6