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3 years ago

Whar is $23.64 rounded to the nearest dollar

2 answers:

7 0

24.00 is the answer to get it so if you look at 23.64 64 is higher then 50 so it goes up to 24.00 but if it was 43 it goes to 23.00. I hope this helped

Scilla [17]3 years ago

5 0

$24 dollars because 64 cents is above 50 cents. So you can round it up.

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10 less than the product of 4 and z

Bumek [7]

Answer:

4z-10

Step-by-step explanation:

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2 years ago

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3 years ago

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Grace [21]

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2 years ago

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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

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