Are the ratios 30:15 and 32:12 equal?

frez [133]

$\left\{\begin{array}{ccc}x+2y=4&\text{multiply both sides by 5}\\4x-5y=42&\text{multiply both sides by 2}\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}5x+10y=20\\8x-10y=84\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad13x=104\qquad\text{divide both sides by 13}\\\\\boxed{x=8}\\\\Answer:\ \boxed{\boxed{x=8}}$

5 0

3 years ago

When people make estimates, they are influenced by anchors to their estimates. A study was conducted in which students were aske

12345 [234]

Answer:

The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0$

where μ1: mean calorie estimation for the cheesecake group and μ2: mean calorie estimation for the organic salad group.

There is enough evidence to support the claim that the mean estimated number of calories in the cheeseburger is lower for the people who thought about the cheesecake first than for the people who thought about the organic fruit salad first (P-value=0.0000002).

Step-by-step explanation:

The question is incomplete:

"Suppose that the study was based on a sample of 20 people who thought about the cheesecake first and 20 people who thought about the organic fruit salad first, and the standard deviation of the number of calories in the cheeseburger was 128 for the people who thought about the cheesecake first and 140 for the people who thought about the organic fruit salad first.

At the 0.01 level of significance, is there evidence that the mean estimated number of calories in the cheeseburger is lower for the people who thought about the cheesecake first than for the people who thought about the organic fruit salad first?"

This is a hypothesis test for the difference between populations means.

The claim is that the mean estimated number of calories in the cheeseburger is lower for the people who thought about the cheesecake first than for the people who thought about the organic fruit salad first.

Then, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0$

The significance level is 0.01.

The sample 1 (cheese cake), of size n1=20 has a mean of 780 and a standard deviation of 128.

The sample 2 (organic salad), of size n2=20 has a mean of 1041 and a standard deviation of 140.

The difference between sample means is Md=-261.

$M_d=M_1-M_2=780-1041=-261$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{128^2}{20}+\dfrac{140^2}{20}}\\\\\\s_{M_d}=\sqrt{819.2+980}=\sqrt{1799.2}=42.417$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{-261-0}{42.417}=\dfrac{-261}{42.417}=-6.153$

The degrees of freedom for this test are:

$df=n_1+n_2-1=20+20-2=38$

This test is a left-tailed test, with 38 degrees of freedom and t=-6.153, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t$

As the P-value (0.0000002) is smaller than the significance level (0.01), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the mean estimated number of calories in the cheeseburger is lower for the people who thought about the cheesecake first than for the people who thought about the organic fruit salad first.

3 0

3 years ago

Answer quickly for 50 points, thank you.

Kaylis [27]

The incorrect answer is B.

All you have to do is plug in x to the formulas and see if the are equal.

4x - 2 = 3x + 6

4(8) - 2 = 3(8) + 6

30 = 30

CORRECT

The second is not only not equal, but it is the wrong kind of formula for what you are trying to prove there.

Hope that helps!

6 0

3 years ago

Read 2 more answers

A store sells boxes of tissue that have 6 tissues in each box. The boxes are arranged 3 across and 4 deep on each shelf. How man

natka813 [3]

Number of boxes = 3 across x 4 deep = 12 total boxes.

Each box has 6 tissues.

Total tissues = 6 per box x 12 boxes = 72 tissues.

4 0

3 years ago

According to survey data, the distribution of arm spans for males is approximately Normal with a mean of 71.4 inches and a stan

Gekata [30.6K]

Answer:

a) 89.97% of men have arm spans between 66 and 76 inches.

b) The z-score for this person's arm span is 5.68. 0% of males have an arm span at least as long as this person

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be 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.

Mean of 71.4 inches and a standard deviation of 3.1 inches.

This means that $\mu = 71.4, \sigma = 3.1$

a. What percentage of men have arm spans between 66 and 76 inches?

The proportion is the pvalue of Z when X = 76 subtracted by the pvalue of Z when X = 66. The percentage is the proportion multiplied by 100.

X = 76

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

$Z = \frac{76 - 71.4}{3.1}$

$Z = 1.48$

$Z = 1.48$ has a pvalue of 0.9306

X = 66

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

$Z = \frac{66 - 71.4}{3.1}$

$Z = -1.74$

$Z = -1.74$ has a pvalue of 0.0409

0.9306 - 0.0409 = 0.8997

0.8997*100% = 89.97%

89.97% of men have arm spans between 66 and 76 inches.

b. A particular professional basketball player has an arm span of almost 89 inches. Find the z-score for this person's arm span. What percentage of males have an arm span at least as long as this person?

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

$Z = \frac{89 - 71.4}{3.1}$

$Z = 5.68$

The z-score for this person's arm span is 5.68.

Z = 5.68 has a pvalue of 1

1 - 1 = 0

0% of males have an arm span at least as long as this person

8 0

3 years ago