All categories

3 years ago

How mucho Does it cost to buy 12 meals if each 3 Meals cost $7.25$ explain please

2 answers:

4 0

$\bf \begin{array}{ccll}meals&\$\\\cline{1-2}3&7.25\\12&x\end{array}\implies \cfrac{3}{12}=\cfrac{7.25}{x}\implies \cfrac{1}{4}=\cfrac{7.25}{x}\\\\\\ x=\cfrac{4\cdot 7.25}{1}\implies x=29$

andrew-mc [135]3 years ago

3 0

So if you have 12 meals and each 3 cost $7.25 then the cost to buy them all is $29.

We can figure this out by dividing each of the meals into groups of 3, and we would get four total groups of 3.

Then we multiply 7.25 by 4 and we get 29, which is the final answer.

You might be interested in

F(3) =-2x^2+9x+5 calculate the following

ElenaW [278]

Answer:

Step-by-step explanation:

2(3)^2+9(3)+5

2(9)+27+5

18+27+5

=50

8 0

3 years ago

A rectangle measures 6 inches by 15 inches. If each dimension of the rectangle is dilated by a scale factor of 1/3 to create a n

Brrunno [24]

I think it is A. I could be wrong

4 0

3 years ago

After a college football team once again lost a game to their archrival, the alumni association conducted a survey to see if alu

valentina_108 [34]

Answer:

P-value for this hypothesis test is 0.00175.

Step-by-step explanation:

We are given that the alumni association conducted a survey to see if alumni were in favor of firing the coach.

A simple random sample of 100 alumni from the population of all living alumni was taken. Sixty-four of the alumni in the sample were in favor of firing the coach.

Let p = proportion of all living alumni who favored firing the coach

SO, Null Hypothesis, $H_0$ : p = 0.50 {means that the majority of alumni are not in favor of firing the coach}

Alternate Hypothesis, $H_A$ : p > 0.50 {means that the majority of alumni are in favor of firing the coach}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of the alumni in the sample who were in favor of firing the coach = $\frac{64}{100}$ = 0.64

n = sample of alumni = 100

So, test statistics = $\frac{0.64-0.50}{{\sqrt{\frac{0.64(1-0.64)}{100} } } } }$

= 2.92

Now, P-value of the hypothesis test is given by ;

P-value = P(Z > 2.92) = 1 - P(Z $\leq$ 2.92)

= 1 - 0.99825 = 0.00175

Therefore, the P-value for this hypothesis test is 0.00175.

4 0

4 years ago

Solve the following system of equations algebraically... NEED HELP PLZ

lana [24]

Answer:

flour $42 , bag of sugar $58

Step-by-step explanation:

let f represent flour and s represent sugar, then

7f + 5s = 584 → (1)

5f + 3s = 384 → (2)

Multiplying (1) by 3 and (2) by - 5 and adding will eliminate s

21f + 15s = 1752 → (3)

- 25f - 15s = - 1920 → (4)

Add (3) and (4) term by term to eliminate s , that is

- 4f = - 168 ( divide both sides by - 4 )

f = 42

Substitute f = 42 into either of the 2 equations and solve for s

Substituting into (1)

7(42) + 5s = 584

294 + 5s = 584 ( subtract 294 from both sides )

5s = 290 ( divide both sides by 5 )

s = 58

The cost of a bag of flour is $42 and cost of sugar is $58

3 0

3 years ago

According to the U.S. Bureau of the Census, the average age of brides marrying for the first time is 23.9 years with a populatio

JulsSmile [24]

Answer:

We conclude that young women are delaying marriage and marrying at a later age.

Step-by-step explanation:

We are given that the average age of brides marrying for the first time is 23.9 years with a population standard deviation of 4.2 years.

The sociologist randomly samples 100 marriage records and determines the average age of the first time brides is 24.9 years.

Let $\mu$ = average age of brides marrying for the first time.

So, Null Hypothesis, $H_0$ : $\mu$ = 23.9 years {means that young women are not delaying marriage and marrying at a later age}

Alternate Hypothesis, $H_A$ : $\mu$ > 23.9 years {means that young women are delaying marriage and marrying at a later age}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age of the first time brides = 24.9 years

$\sigma$ = population standard deviation = 4.2 years

n = sample of marriage records = 100

So, the test statistics = $\frac{24.9-23.9}{\frac{4.2}{\sqrt{100} } }$

= 2.381

The value of z test statistics is 2.381.

Now, at 1% significance level the z table gives critical value of and 2.326 for right-tailed test.

Since our test statistic is more than the critical value of z as 2.381 > 2.326, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that young women are delaying marriage and marrying at a later age.

5 0

3 years ago