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shepuryov [24]
3 years ago
5

46 qt = ?? ozPLEASEE HELP ASAP

Mathematics
2 answers:
Dennis_Churaev [7]3 years ago
8 0

Answer:

1472 oz

Step-by-step explanation:

recall that there are 32 oz in a quart

i.e 1 quart ----> 32 oz

46 quarts ----> 32 x 46 = 1472 oz

Hoochie [10]3 years ago
3 0

Answer:

1472

Step-by-step explanation:

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Answer:

This is false statement.

Step-by-step explanation:

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⇒ 1 hectogram = 100 gram

⇒ 0.01 gram = 0.0001 hectogram

Thus the given statement is false.

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The World Issues club has decided to donate 60% of all their fundraising activities this year to Stephen Lewis
andreyandreev [35.5K]

Answer:

a. Let the variable be x for the fundraising activities and M as the revenue for foundation.

b. M =0.60x

c. $43.2

d. $1416.67

Step-by-step explanation:

Given that:

The World Issues club donates 60% of the total of their fundraising activities.

Answer a.

Let us choose the variable x to represent the money earned during fundraising activities and M for the revenue generated for foundation.

Answer b.

Foundation will receive 60% of the total of the fundraising activities.

Equation to determine the money that will be received by foundation:

M = 60\%\ of\ x\\OR\\M = 0.6x

Answer c.

Given that x = $72, M = ?

Putting the value of x in the equation above:

M = 0.6 \times 72\\\Rightarrow \$43.2

Answer d.

Given that M = $850, x = ?

Putting the value of M in the equation above to find x:

850= 0.6 \times x\\\Rightarrow x = \dfrac{850}{0.6}\\\Rightarrow x = \$ 1416.67

So, the answers are:

a. Let the variable be x for the fundraising activities and M as the revenue for foundation.

b. M =0.60x

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5 0
3 years ago
Does going to a private university increase the chance that a student will graduate with student loan debt? A national poll by t
Snowcat [4.5K]

Answer:

Null hypothesis:p \leq 0.69  

Alternative hypothesis:p > 0.69  

Step-by-step explanation:

1) Data given and notation

n=1500 represent the random sample taken

X represent the number of graduates that had student loan debt in 2014

\hat p=0.71 estimated proportion of adults that said that it is morally wrong to not report all income on tax returns

p_o=0.69 is the value that we want to test

\alpha represent the significance level  

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that if there was a significant increase in the proportion of student loan debt for public and nonprofit colleges in 2014 respect to the value of 2013.:  

Null hypothesis:p \leq 0.69  

Alternative hypothesis:p > 0.69  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.71 -0.69}{\sqrt{\frac{0.69(1-0.69)}{1500}}}=1.675  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed is \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(Z>1.675)=0.047  

If we compare the p value obtained and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults that said that it is morally wrong to not report all income on tax returns  is not significantly higher than 0.69.  

5 0
3 years ago
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