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2 years ago

I need help with this

Mathematics

1 answer:

adelina 88 [10]2 years ago

7 0

Answer:

a. 0.444444

b. 0.87

Step-by-step explanation:

a. (-2).(-4.5)

= -2 / -4.5

= 0.44444

u will do the same to the second question

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the library is trying to decide which book to feature in next month's young adult book club, salt to the sea or the sun is also

kozerog [31]

Answer:

Salt to the Sea.

Step-by-step explanation:

The Sun is also a Star is a romance/Love story, while Salt to the Sea is a young adult book about four individuals in World War II.

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3 years ago

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A leaky pool loses 2 and one-fourth inches of water each day. Jermaine and Mildred are asked to determine the change in the amou

Licemer1 [7]

Answer:

Mildred is correct because she distributed 5 correctly

Step-by-step explanation:

We cannot tell exactly which solution goes with which student, but we believe you are saying ...

Mildred’s Solution:

5 (Negative 2 and one-fourth) = 5 (negative 2 minus one-fourth)

= negative 10 minus StartFraction 5 over 4 EndFraction

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This solution describes the correct distribution of 5. Whoever it belongs to gave the correct solution.

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4 years ago

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Line z has a slope of 2.5 and goes through point (4, 13). Which equation best represents line z?

tatuchka [14]

Answer:

y = my + b

Step-by-step explanation:

According to my calculations. the equation best representing Z is the one above

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3 years ago

Eduardwww [97]

The answer to the question is b

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4 years ago

Two alloys A and B are being used to manufacture a certain steel product. An experiment needs to be designed to compare the two

sveta [45]

Answer:

Step-by-step explanation:

From the given information, we can compute the table showing the summarized statistics of the two alloys A & B:

Alloy A Alloy B

Sample mean $\bar {x} _A = 49.5$ $\bar {x} _B = 45.5$

Equal standard deviation $\sigma_A = 5$ $\sigma_B= 5$

Sample size $n_A = 30$ $n_{B}= 30$

Mean of the sampling distribution is :

$\mu_{\bar{X_1}-\bar{X_2}}= 49.5-45.5 \\ \\ = 4.0$

Standard deviation of sampling distribution:

$\sigma_{\bar{x_1}-\bar{x_2} }= \sqrt{\sigma^2_{\bar{x_1}-\bar{x_2} }} \\ \\ = \sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma^2_2}{n_2} } \\ \\ =\sqrt{\dfrac{25}{30}+\dfrac{25}{30}} \\\\=\sqrt{1.667} \\ \\ =1.2909$

Hypothesis testing.

Null hypothesis: $H_o: \mu_A -\mu_B = 0$

Alternative hypothesis: $H_A:\mu_A -\mu_B > 4$

The required probability is:

$P(\overline X_A - \overline X_B>4|\mu_A - \mu_B) = P\Big (\dfrac{(\overline X_A - \overline X_B)-\mu_{X_A-X_B}}{\sigma_{\overline x_A -\overline x_B}} > \dfrac{4 - \mu_{X_A-\overline X_B}}{\sigma _{\overline x_A - \overline X_B}} \Big) \\ \\ = P \Big( z > \dfrac{4-0}{1.2909}\Big) \\ \\ = P(z \ge 3.10)\\ \\ = 1 - P(z < 3.10) \\ \\ \text{Using EXCEL Function:} \\ \\ = 1 - [NORMDIST(3.10)] \\ \\ = 1- 0.999032 \\ \\ 0.000968 \\ \\ \simeq 0.0010$

This implies that a minimal chance of probability shows that the difference of 4 is not likely, provided that the two population means are the same.

Since the P-value is very small which is lower than any level of significance.

Then, we reject $H_o$ and conclude that there is enough evidence to fully support alloy A.

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3 years ago