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2 years ago

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Jerry’s rhombus has a base of 15 inches and a height of 9 inches. Charlene’s rhombus has a base and a height that are 13 the bas

e and height of Jerry’s rhombus. Compare the area of Charlene’s rhombus to the area of Jerry’s rhombus.

2 answers:

neonofarm [45]2 years ago

7 0

Answer: The area of Charlene's rhombus is nine times smaller than the area of Jerry's rhombus.

Step-by-step explanation:

I will assume that the exercise says " $\frac{1}{3}$ times the base and height of Jerry’s rhombus".

The area of a rhombus can be calculated with the following formula:

$A=b*a$

Where "b" is the length of the base and "a" is the altitude or the height.

Then, you can calculate the area using the formula shown above.

Therefore, you get:

1. Jerry's rhombus:

$A_1=(15\ in)(9\ in)\\\\A_1=135\ in^2$

2. Charlene's rhombus:

$A_2=(\frac{1}{3}*15\ in)(\frac{1}{3}*9\ in)\\\\A_2=(5\ in)(3\ in)\\\\A_2=15\ in^2$

Dividing the area calculated, you get:

$\frac{135\ in^2}{15\ in^2}=9$

Therefore, you can conclude that the area of Charlene's rhombus is nine times smaller than the area of Jerry's rhombus.

Eddi Din [679]2 years ago

6 0

Answer:

Step-by-step explanation:

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Answer: (80% , 85%)

Step-by-step explanation:

We know that, confidence interval for population proportion is given by :-

(p-E , p+E)

, where p = sample proportion , E = Margin of error.

Given : Proportion of elementary school teachers who are female = 82%.

The article also states the maximum error of their estimate = 3%.

Then, the 90% confidence interval for the proportion of elementary school teachers who are female will be :

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= (80% , 85%)

Hence, the resulting 90% confidence interval for the proportion of elementary school teachers who are female = (80% , 85%)

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2 years ago

Suppose X is the breaking strength (newtons) of a material, and X is normally distributed with

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Answer:

a) 0.997 is the probability that the breaking strength is at least 772 newtons.

b) 0.974 is the probability that this material has a breaking strength of at least 772 but not more than 820

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 800 newtons

Standard Deviation, σ = 10 newtons

We are given that the distribution of breaking strength is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P( breaking strength of at least 772 newtons)

$P(x \geq 772)$

$P( x \geq 772) = P( z \geq \displaystyle\frac{772 - 800}{10}) = P(z \geq -2.8)$

$= 1 - P(z$

Calculation the value from standard normal z table, we have,

$P(x \geq 772) = 1 - 0.003 = 0.997 = 99.7\%$

0.997 is the probability that the breaking strength is at least 772 newtons.

b) P( breaking strength of at least 772 but not more than 820)

$P(772 \leq x \leq 820) = P(\displaystyle\frac{772 - 800}{10} \leq z \leq \displaystyle\frac{820-800}{10}) = P(-2.8 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2.8)\\= 0.977 - 0.003 = 0.974 = 97.4\%$

$P(772 \leq x \leq 820) = 97.4\\%$

0.974 is the probability that this material has a breaking strength of at least 772 but not more than 820.

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2 years ago

A local arts council has 200 members. The council president wanted to estimate the percent of its members who have had experienc

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Answer:

Yes the sample can be use to make inference

Step-by-step explanation:

The inference is possible if the conditions:

p*n > 10 and q*n > 10

where p and q are the proportion probability of success and q = 1 - p

n is sample size

Then p = 12 / 30 = 0,4 q = 1 - 0,4 q = 0,6

And p*n = 0,4 * 30 = 12 12 > 10

And q*n = 0,6 * 30 = 18 18 > 10

Therefore with that sample the conditions to approximate the binomial distribution to a Normal distribution are met

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2 years ago

Which ordered pair minimizes the objective function C = 60x + 85y?. (0, 160). (55, 70). (80, 50). (170, 0)

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Step-by-step explanation:

The given objective function is $C = 60x + 85y$

To check that which ordered pair is minimizing the objective function-we will substitute all the ordered pair one by one in the given function and we will calculate the value of C. The ordered pair which will give the minimum value will be the answer.

For (0, 160)

$C=60\times 0+85\times 160=13600$

For (55, 70)

$C=60\times 55+85\times 70=9250$

For (80, 50)

$C=60\times 80+85\times 50=9050$

For (80, 50)

$C=60\times 170+85\times 0=10200$

Hence the ordered pair which minimizes the objective function is (80, 50)

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We have been given that Melissa owns a credit card with a 8.9% APR. The balance after her last billing cycle was $550.

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The finance charge would be 0.74166% of $550.

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