For the data in the table, does y vary directly with x? If it does, write an equation for the direct variation.

A triangle with area 45 square inches has a height that is two less than four times the width.

olga2289 [7]

Answer:base=5

Height =18

Step-by-step explanation:

18×5=90

5 0

3 years ago

Read 2 more answers

The total amount of fence around a rectangular play area must be at least 90 feet, but no more than 120 feet. If the length of t

sergejj [24]

The perimeter of a triangle has the formula 2l + 2w. Since the length is twice as long as the width, l = 2w. We use this equations to two possible perimeters, 90 and 120 feet.

P = 90 feet
90 = 2(2w) + 2w
90 = 6w
w = 15 feet

P = 120 feet
120 = 2(2w) + 2w
120 = 6w
w = 20

Thus, the range of w is from 15 to 20 feet. The answer is <span>15 ≤ w ≤ 20</span>

6 0

3 years ago

The Cooking Club made some pies to sell

Alenkinab [10]

The club made 5 pies. If each pie is cut into 5 slices, and there are now 40 slices after the cafeteria donated, divide 40 by 5, which is 8. Subtract the 3 that the cafeteria donated.

3 0

3 years ago

40%40% of the memberships sold by Gym A were gold memberships. 25%25% of the memberships sold by Gym B were gold memberships. Wh

Darya [45]

Where are the statements?
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5 0

3 years ago

The mean water temperature downstream from a power plant coolingtower discharge pipe should be no more than 102oF. Pastexperienc

nata0808 [166]

Answer:

(a) Yes, there is evidence that the water temperature is acceptable at $\alpha = 0.05$ (b) 0.9987 (c) 6.647274e-06

Step-by-step explanation:

Let X be the random variable that represents the water temperature. The water temperature has been measured on n = 9 randomly chosen days (a small sample), the sample average temperature is $\bar{x}$ = 100°F and $\sigma$ = 2°F. We suppose that X is normally distributed.

We have the following null and alternative hypothesis

$H_{0}: \mu = 102$ vs $H_{1}: \mu > 102$ (upper-tail alternative)

We will use the test statistic

$Z = \frac{\bar{X}-102}{2/\sqrt{9}}$ and the observed value is

$z_{0} = \frac{100-102}{2/\sqrt{9}} = -3$ .

(a) The rejection region is given by RR = {z | z > 1.6448} where 1.6448 is the 95th quantile of the standard normal distribution. Because the observed value -3 does not belong to RR, we fail to reject the null hypothesis. In other words, there is evidence that the water temperature is acceptable at $\alpha = 0.05$ .

(b) The p-value for this test is given by P(Z > -3) = 0.9987

(c) P(Accepting $H_{0}$ when $\mu = 106$ ) = P(The observed value is not in RR when $\mu = 106$ ) = P( $\frac{\bar{X}-102}{2/\sqrt{9}}$ < 1.6448 when $\mu = 106$ ) = P( $\bar{X}$ < 102 + (1.6448)( $2/\sqrt{9}$ ) when $\mu = 106$ ) = P( $\bar{X}$ < 103.0965) when $\mu = 106$ ) = P( $(\bar{X}-106)/(2/\sqrt{9}) < (103.0965-106)/(2/\sqrt{9})$ )) = P(Z < -4.3552) = 6.647274e-06

7 0

3 years ago