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2 years ago

Please help me.

Mathematics

2 answers:

Keith_Richards [23]2 years ago

8 0

Answer:

$180

Step-by-step explanation:

Since the discount is 40 percent and is also $72, We will cross multiply. 72 x 100 = 40 x what?

7200/40 = 180

The mirror was originally $180.

AURORKA [14]2 years ago

8 0

$180 :))) hope this helps

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Un jardín rectangular de 50 cm de largo por 34 m de ancho está rodeado por un camino de arena uniforme.Halla la anchura de dicho

Anastasy [175]

Answer:

5.85 m

Step-by-step explanation:

The width of the sand road can be calculated knowing its area and the dimensions of the rectangular garden as follows:

$A_{g} = a.b$

Where:

Ag: is the area of the rectangular garden

a: is the length of the rectangular garden = 50 cm = 0.5 m

b: is the width of the rectangular garden = 34 m

$A_{s} = 540 m^{2}$

Where:

As: is the area of the sand road

The relation between the area of the sand road and the area of the rectangular garden is the following:

$A_{s} + A_{g} = (a+2x)*(b+2x)$

$540 m^{2} + 0.5m*34m = ab + 2ax + 2bx + 4x^{2}$

$557 m^{2} = ab + 2x(a + b) + 4x^{2}$

$557 m^{2} - 17m^{2} - 2x(34.5 m) - 4x^{2} = 0$

$540 - 69x - 4x^{2} = 0$

By solving the above equation for x we have two solutions:

x₁ = -23.10 m

x₂ = 5.85 m

Taking the positive value, we have that the width of the sand road is 5.85 m.

I hope it helps you!

7 0

3 years ago

The residual plot for a data set is shown.

BaLLatris [955]

The correct answer is: D. The regression line is not a good model because the points in the residual plot form a curve. (i know this because i just passed the test ;) )

5 0

3 years ago

Read 2 more answers

Describe the sampling distribution of p(hat). Assume the size of the population is 30,000.

Naya [18.7K]

Answer:

a) $\mathbf{\mu_ \hat p = 0.6}$

b) $\mathbf{\sigma_p =0.01732}$

Step-by-step explanation:

Given that:

population mean $\mu$ = 30,000

sample size n = 800

population proportion p = 0.6

The mean of the the sampling distribution is equal to the population proportion.

$\mu_ \hat p = p$

$\mathbf{\mu_ \hat p = 0.6}$

The standard deviation of the sampling distribution can be estimated by using the formula:

$\sigma_p = \sqrt{\dfrac{p(1-p)}{n}}$

$\sigma_p = \sqrt{\dfrac{0.6(1-0.6)}{800}}$

$\sigma_p = \sqrt{\dfrac{0.6(0.4)}{800}}$

$\sigma_p = \sqrt{\dfrac{0.24}{800}}$

$\sigma_p = \sqrt{3 \times 10^{-4}}$

$\mathbf{\sigma_p =0.01732}$

7 0

3 years ago

explain why Shirley Chisholm was a political pioneer. __________________________________________________________________________

irga5000 [103]

Answer: She had already surprised everyone by becoming the first black woman in Congress after an upset victory in 1968. Then Shirley Chisholm signed up for work as a census taker in Brooklyn, where she represented a range of struggling neighborhoods.

It was a thankless task; many of the “enumerators” for the 1970 census quit because so many poor black and Hispanic residents refused to answer questions or even open the door.

Their distrust in government ran deep, The Times reported, with some fearing that giving up their personal information would lead to genocide.

Ms. Chisholm, a daughter of immigrants from Barbados who studied American history with the zeal of a woman determined to shape it, understood such sentiments. She also embodied what was needed to bring those New Yorkers into the fold. It wasn’t pontificating. It wasn’t condescending, or scolding; it required the same charm and resolve she showed first as an educator, then as a politician.

“I do not see myself as a lawmaker, an innovator in the field of legislation,” she wrote in her 1970 autobiography, “Unbought and Unbossed.” “America has the laws and the material resources it takes to insure justice for all its people. What it lacks is the heart, the humanity, the Christian love that it would take.”

8 0

3 years ago

Suppose the horses in a large stable have a mean weight of 975lbs, and a standard deviation of 52lbs. What is the probability th

Lubov Fominskaja [6]

Answer:

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 975, \sigma = 52, n = 31, s = \frac{52}{\sqrt{31}} = 9.34$

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable?

pvalue of Z when X = 975 + 15 = 990 subtracted by the pvalue of Z when X = 975 - 15 = 960. So

X = 990

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{990 - 975}{9.34}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463

X = 960

$Z = \frac{X - \mu}{s}$

$Z = \frac{960 - 975}{9.34}$

$Z = -1.61$

$Z = -1.61$ has a pvalue of 0.0537

0.9463 - 0.0537 = 0.8926

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

7 0

3 years ago