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3 years ago

Find the interest is you invest $3500 for 9 years at a rate of 4.5%

Mathematics

1 answer:

dybincka [34]3 years ago

5 0

Aproxx. 90 or 89 it was to long to type so im giving you the estiment

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Find the value of the lesser root of x2 - 7x + 12 = 0.<br> A) -3 <br> B) -1 <br> C) 1 <br> D) 3

kvv77 [185]

The answer to your question is D) 3

8 0

3 years ago

Read 2 more answers

Lifetime of $1 Bills The average lifetime of circulated $1 bills is 18 months. A researcher believes that the average lifetime i

OLEGan [10]

<h2>Answer with explanation:</h2>

Let $\mu$ be the population mean lifetime of circulated $1 bills.

By considering the given information , we have :-

$H_0:\mu=18\\\\H_a:\mu\neq18$

Since the alternative hypotheses is two tailed so the test is a two tailed test.

We assume that the lifetime of circulated $1 bills is normally distributed.

Given : Sample size : n=50 , which is greater than 30 .

It means the sample is large so we use z-test.

Sample mean : $\overline{x}=18.8$

Standard deviation : $\sigma=2.8$

Test statistic for population mean :-

$z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

$\Rightarrow\ z=\dfrac{18.8-18}{\dfrac{2.8}{\sqrt{50}}}\approx2.02$

The p-value= $2P(z>2.02)=0.0433834$

Since the p-value (0.0433834) is greater than the significance level (0.02) , so we do not reject the null hypothesis.

Hence, we conclude that we do not have enough evidence to support the alternative hypothesis that the average lifetime of a circulated $1 bill differs from 18 months.

6 0

3 years ago

Solve the equation and check your solution. (If all real numbers are solutions, enter REALS. If there is no solution, enter NO S

shepuryov [24]

Answer:

y=6/5 or 1 1/5

Step-by-step explanation:

combine like terms

5y+1= 10y-5

subtract 5y from both sides

1=5y-5

add 5 to both sides

6=5y

divide by 5 on both sides

6/5=y

5 0

3 years ago

What is the supplement measurement of a 115 degree angle?

erma4kov [3.2K]

Let the supplement be s.

180=s+115 => s=180-115= 65 degrees

8 0

3 years ago

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48% of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually

oksian1 [2.3K]

Answer:

There is no sufficient evidence to support the executive claim

Step-by-step explanation:

From the question we are told that

The population proportion is $p = 0.48$

The sample proportion is $\r p = 0.45$

The sample size is $n = 300$

The level of significance is $\alpha = 0.02$

The null hypothesis is $H_o : p= 0.48$

The alternative hypothesis is $H_a : p \ne 0.48$

Generally the test statistics is mathematically evaluated as

$t = \frac{\r p - p }{ \sqrt{ \frac{p(1 - p )}{n} } }$

=> $t = \frac{0.45 - 0.48 }{ \sqrt{ \frac{0.48 (1 - 0.48 )}{300} } }$

=> $t = -1.04$

The p-value is mathematically represented as

$p-value = 2P(z > |-1.04|)$

Form the z-table

$P(z > |-1.04|) = 0.15$

=> $p-value = 2 * 0.15$

=> $p-value = 0.3$

Given that $p-value > \alpha$ we fail to reject the null hypothesis

Hence we can conclude that there is no sufficient evidence to support the executive claim

5 0

3 years ago