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3 years ago

Using the equation, y=10000(0.97)x predict the purchasing power of $10000 ten years later

2 answers:

8 0

Using the equation, y= 10,000(0.97)^ , predict the purchasing power of $10,000 ten years later. 
$7374

Andrew [12]3 years ago

7 0

Answer:

The purchasing power is $7374

Step-by-step explanation:

We are given an equation $y=10000(0.97)^x$

where,

y represents purchasing power.

x represents the number of years.

We need to find purchasing power of $10000 after 10 years.

Put x=10 into equation and solve for y

$y=10000(0.97)^{10}$

$y=7374.24\approx 7374$

After 10 years the purchasing power of $10000 becomes $7374

As we can see the value of the product decrease.

Hence, The purchasing power is $7374

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The mean resting pulse rate for men is 72 beats per minute. A simple random sample of men who regularly work out at Mitch's Gym

BARSIC [14]

Answer:

We conclude that the mean resting pulse rate for men is 72 beats per minute.

Step-by-step explanation:

We are given that he mean resting pulse rate for men is 72 beats per minute. Assume that the standard deviation of the resting pulse rates of all men who work out at Mitch's Gym is known to be 2.6 beats per minute.

A simple random sample of men who regularly work out at Mitch's Gym is obtained and their resting pulse rates (in beats per minute) are listed below;

87, 89, 69, 63, 70, 65, 88, 84, 58, 53, 66, 70.

Let $\mu$ = mean resting pulse rate for men.

SO, Null Hypothesis, $H_0$ : $\mu$ = 72 beats/minute {means that the mean resting pulse rate for men is 72 beats per minute}

Alternate Hypothesis, $H_A$ : $\mu$ < 72 beats/minute {means that the mean resting pulse rate for men is less than 72 beats per minute}

The test statistics that would be used here One-sample z-test statistics as we know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean mean resting pulse rate = $\frac{\sum X}{n}$

= $\frac{87+ 89+ 69 +63 +70 +65+ 88+ 84+ 58+ 53+ 66+ 70}{12}$ = 71.83 beats/minute

$\sigma$ = population standard deviation = 2.6 beats per minute

n = sample of men = 12

So, the test statistics = $\frac{71.83 - 72}{\frac{2.6}{\sqrt{12} } }$

= -0.23

The value of z test statistics is -0.23.

Also, P-value of the test statistics is given by;

P(Z < -0.23) = 1 - P(Z $\leq$ 0.23)

= 1 - 0.59095 = 0.40905

Now, at 0.05 significance level the z table gives critical value of -1.645 for left-tailed test.

Since our test statistic is more than the critical value of z as -0.23 > -1.645, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean resting pulse rate for men is 72 beats per minute.

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3 years ago

0 = 2x2 - 6x + 4 What are the solutions to

Ainat [17]

The answer would be .5

6 0

3 years ago

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Renting a porch for the day there is a $50. Fee plus .50 per mile with a budget of $100. How many miles can you go

UkoKoshka [18]

Answer:2