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Rom4ik [11]
3 years ago
15

Irene tutors fellow students to earn spending money. A designer purse she wants to buy is $170. She has already earned and set a

side $51 for the purse. If she charges $17 per hour of tutoring, how many hours (h) must she tutor to have at least enough money to buy the purse? A. H > 10 B. H > 7 C. H > 3 D. H > 13
Mathematics
1 answer:
zimovet [89]3 years ago
4 0

Answer:

A

Step-by-step explanation:

$170 - $51 = $119 more needed for the purse

she earns $17 per hour so

$119 ÷ $17 = 7

she should tutor 7 hours so h>10

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Oksana_A [137]

Answer:

Step-by-step explanation: You have to do the distributive property to get the answer.

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Can anyone help me with 15a b c
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3 years ago
A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro
-Dominant- [34]

Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

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answer

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step-by-step explanation

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3 years ago
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Answer:

A: $0.25

Step-by-step explanation:

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