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siniylev [52]
3 years ago
9

Need this ASAP!! Would help if you could answer as quick as possible x

Mathematics
2 answers:
zaharov [31]3 years ago
7 0
Question: -7(3p + c)

First you need to multiply the () by -7 or in other words distribute. In order to do this multiply each value in the () by -7

-7(3p) = -21p
-7(c) = -7c

Now rewrite your equation

-21p-7c

There you go!
Ronch [10]3 years ago
6 0

Answer:

-21p - 7c

Step-by-step explanation:

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

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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

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b)

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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