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Cerrena [4.2K]
3 years ago
12

Help me please I need help on this question!!!!!

Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
5 0
The answer is b I think
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Julli [10]

Answer:

7

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
2. The number of nickels that Christine has is
aev [14]

Answer:

The number of nickels is  15

The number of Dimes is 3

Step-by-step explanation:

Let the number of dimes  Christine has be d

The number of nickels with christine be n

Then the number of Nickels  Christine has will be  

n = 5d ----------------------------------(1)

Now The total value of both dimes and nickels is  = $1.05

This can be represented as

5n +10d = 105  (in cents)

From eq(1)

5(5d) + 10d = 105

25d+10d = 105

35d = 105

d = \frac{105}{35}

d = 3----------------------------------(2)

Substituting (2) in (1), we get

n = 5(3)

n = 15

7 0
3 years ago
There are 1800 students in a school, if the ratio of teachers to students is 2: 7. How many teachers are at the school?
Vinvika [58]

Answer: 258

Step-by-step explanation:

7/9 = 1800

2/9 = ?

= 258

8 0
3 years ago
According to a Pew Research Center study, in May 2011, 35% of all American adults had a smart phone (one which the user can use
taurus [48]

Answer:

z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82  

p_v =P(z>1.82)=0.034  

So the p value obtained was a very low value and using the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis

And the best conclusion would be:

There is enough evidence to show that more than 35% of community college students own a smart phone (Pvalue = 0.034).

And for the second case the correct system of hypothesis is:

H0: p = 0.078; Ha: p > 0.078

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

\hat p=\frac{120}{300}=0.4 estimated proportion of college students that have a smart phone

p_o=0.35 is the value that we want to test

\alpha=0.1 represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is >0.35.:  

Null hypothesis:p\leq 0.35  

Alternative hypothesis:p > 0.35  

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{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82  

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=0.1. 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>1.82)=0.034  

So the p value obtained was a very low value and using the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis

And the best conclusion would be:

There is enough evidence to show that more than 35% of community college students own a smart phone (Pvalue = 0.034).

And for the second case the correct system of hypothesis is:

H0: p = 0.078; Ha: p > 0.078

4 0
3 years ago
Sam’s fish tank has a base area of 187 in2 and a height of 16
inessss [21]

997 1/3 in 3

hope it helps

fhcdfyxvbonmjcf

7 0
3 years ago
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