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Margarita [4]
2 years ago
5

PLEASE HELP!! NO LINKS!! JUST ANSWER THE QUESTIONS BELOW THE WORD “QUESTIONS:”

Mathematics
1 answer:
Otrada [13]2 years ago
4 0

Answer:

maybe is 8 l'm not sure

Step-by-step explanation:

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A man spent 15/16 of his entire fortune in buying a car $7500. How much money did he possess?
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X=his fortune
15/16X=7500
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Dylan owes half as much money as he used to owe. If he used to owe $28, which of the following expressions would reflect how muc
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Fill in the blanks
brilliants [131]

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1. 100 which Is the easier one

2. 5

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2 years ago
2. What is the effective annual rate of an investment that pays 6% for 5 years, compounded semiannually?
Ede4ka [16]

Answer:

effective annual rate is 6.16 %

Step-by-step explanation:

given data

rate = 6 % = 0.06

time 5 year = 10 semi annually

to find out

effective annual rate

solution

we know formula for annual effective rate of interest is

rate of  interest = (1+ r/n)^{n} -1

put here all value

rate of interest = (1+ 0.06/10)^{10} -1

rate of  interest = (1+ 0.06/10)^{10} -1

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8 0
3 years ago
A group of 54 college students from a certain liberal arts college were randomly sampled and asked about the number of alcoholic
Ket [755]

Answer:

t=\frac{5.25-4.73}{\frac{3.98}{\sqrt{54}}}=0.9601  

P-value  

First we find the degrees of freedom given by:

df = n-1= 54-1=53

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

p_v =2*P(t_{53}>0.9601)=0.3414  

Step-by-step explanation:

Data given and notation  

\bar X=5.25 represent the sample mean

s=3.98 represent the sample standard deviation

n=54 sample size  

\mu_o =4.73 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z 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 mean differs from 4.73, the system of hypothesis would be:  

Null hypothesis:\mu =4.73  

Alternative hypothesis:\mu \neq 4.73  

Since we know don't the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

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

t=\frac{5.25-4.73}{\frac{3.98}{\sqrt{54}}}=0.9601  

P-value  

First we find the degrees of freedom given by:

df = n-1= 54-1=53

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

p_v =2*P(t_{53}>0.9601)=0.3414  

7 0
3 years ago
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