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ArbitrLikvidat [17]
3 years ago
15

An airplane flies 2,951 miles in 6.5 hours. What is the speed of the airplane in miles per hour?

Mathematics
1 answer:
Alex_Xolod [135]3 years ago
4 0
The answer is 454. How to solve? Simple! Just divide 2951 and 6.5.
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At Thanksgiving, Mrs. Jones overcooked the turkey so all but 5 of it had to be thrown away. The 8 people at dinner that
VashaNatasha [74]

<em>The question has inconsistent or incomplete data, so I'm filling the holes with key data.</em>

Answer:

<em>Every people at dinner received one-tenth of the original turkey= 0.1</em>

Step-by-step explanation:

<u>Proportions</u>

If some fraction a/b of a whole total M is to be computed and later removed, we proceed as follows

* Compute the portion to be removed as a/b*M

* Subtract it from the total quantity: M-a/b*M=M(1-a/b)

I'm assuming 1/5 of the turkey was lost due to overcooking. It means that (1-1/5) of the turkey remained for dinner, that is, 4/5 of the turkey.

Each people at dinner received the same amount of the remaining, so we must divide 4/5 by 8, to get 4/40, or 1/10. It means that every people at dinner received one-tenth of the original turkey

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3 years ago
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The solutions of x2=169
tresset_1 [31]

Answer:

x=13

Step-by-step explanation:

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I need help with this plz bro. Plz
Bogdan [553]

Answer:

I only got B

Step-by-step explanation:

Put the x-values into the equation and solve for each.

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Write the following decimal number in its equivalent fraction form. Show all work for full credit.
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Negative 3143 over 1000 divide -3.143 by 1000 and there’s your answer
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3 years ago
Ten years ago 53% of American families owned stocks or stock funds. Sample data collected by the Investment Company Institute in
Alborosie

Answer:

a) Null hypothesis:p\geq 0.53  

Alternative hypothesis:p < 0.53  

b) z=\frac{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429  

p_v =P(Z

c) So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .  

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

\hat p=0.46 estimated proportion of American families owning stocks or stock funds

p_o=0.53 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

Part a

We need to conduct a hypothesis in order to test the claim that proportion is less than 0.53 or 53%.:  

Null hypothesis:p\geq 0.53  

Alternative hypothesis:p < 0.53  

Part b

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.

Calculate the statistic  

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

z=\frac{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429  

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 provided \alpha=0.01. The next step would be calculate the p value for this test.  

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

p_v =P(Z

Part c  

So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .  

7 0
2 years ago
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