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3 years ago

Find the least whole number that can replace ? To make the statement true ? ÷9>700

Mathematics

1 answer:

Elza [17]3 years ago

3 0

Answer:

6301.

Step-by-step explanation:

First we find the integer such that ? / 9 = 700.

? = 700 * 9 = 6300.

Therefore the least whole number for ? / 9 > 700 should be 6301.

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Which parent function is graphed below

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Answer:

$y = {x}^{3}$

Cubic Function

Put it into a graphing calculator like Desmos or TI-84

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A physical education department wanted a single test of upper body strength that was easy to administer. Dips on the parallel ba

Gennadij [26K]

Answer:

24.23 ; 13.47 ; 0.782 ; p value < 0.00001 ;7.45

Step-by-step explanation:

Given that :

ΣX = 3,416, ΣY = 1,899, SDX = 8.84, SDY = 4.70, ΣZXZY = 109.416

Sample size, N = 141

Mean of X:

£X / N = 3416 / 141 = 24.23

Mean of Y:

£Y / N = 1899 / 141 = 13.47

B.) Correlation between the 2 variables :

£ZXZY / N - 1

= 109.416 / 140

= 0.782

C.)

Level of significance and p value :

T = [Rsqrt(n-2)] ÷ sqrt(1 - R²)

T = [0.782 * sqrt(139)] ÷ sqrt(1-0.782^2)

T = 14.79

P value from Tscore calculator

Df = 141 - 2 = 139 ; α = 0.05

Pvalue = 0.00001

Pvalue < α ; Hence, a significant positive relationship exists.

To obtain regression model :

Slope = £ZXZY / SDX²

m = 109.416 / 8.84^2 = 1. 4

yintercept = meanY - slope*x

yintercept = 13.47 - 1.4*24.3

yintercept = - 20.55

Equation in slope intercept form:

y = mx + c

y = 1.4x - 20.55

For x = 20

y = 1.4(20) - 20.55

y = 7.45

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3 years ago

In a sample of 234 individuals over the age of 25, chosen at random from the state of Oregon, 48 did not have a high school dipl

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Answer:

The upper bound of a 95% confidence interval for the proportion of individuals over the age of 25 in Oregon who do not have a high school diploma is of 0.2568.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

In a sample of 234 individuals over the age of 25, chosen at random from the state of Oregon, 48 did not have a high school diploma.

This means that $n = 234, \pi = \frac{48}{234} = 0.2051$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2051 + 1.96\sqrt{\frac{0.2051*0.7949}{234}} = 0.2568$

The upper bound of a 95% confidence interval for the proportion of individuals over the age of 25 in Oregon who do not have a high school diploma is of 0.2568.

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3 years ago

Chad will need at least 24 minutes to complete the 5K race. However, he wants to finish in under 30 minutes.

dedylja [7]

Answer:

The required inequality is: $24 \leq x$

The graph is shown in figure.

Step-by-step explanation:

Consider the provided information.

Chad will need at least 24 minutes to complete the 5K race. However, he wants to finish in under 30 minutes.

Let the time taken to finish the race is represented by x.

Chad need at least 24 minutes and he wants to finish in under 30 minutes.

Thus, the required inequality is:

$24 \leq x$

Now draw the graph of the inequality.

Use a dot or close circle to represents ≤ and use an open circle to represent <.

The value of x is greater than or equal to 24 but less than 30, Thus the required graph is shown in figure.

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anygoal [31]

Answer:

Step-by-step explanation:

The answer is B, 18

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