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A lamp pole with a height of 10 ft casts a shadow of 3 ft. What is the height of a building that casts a shadow of 15 ft at the

seraphim [82]

Answer:

4.5

Step-by-step explanation:

3 0

3 years ago

Find an equation of a line through (0, -7) and parallel to the line y=1/2x+5.

marysya [2.9K]

Answer:

The answer to your question is y = 1/2x - 7

Step-by-step explanation:

Data

Point = (0, -7)

line = y = 1/2x + 5

Parallel line have the same slope.

Slope of the first line = 1/2

Slope of the new line = 1/2

Equation of the line

y - y1 = m(x - x1)

Substitution

y + 7 = 1/2(x + 0)

Simplification

y + 7 = 1/2x

result

y = 1/2x - 7

5 0

4 years ago

Sam had $234.15 in his bank account and wrote a check for $56.29. His Grandmother then gave him $45 to deposit into his account.

Sauron [17]

Answer:

the answer is $222.86

Step-by-step explanation:

you would take the 234.15 and subtract the 56.29 dollars and then you would add the 45 dollars to the account to equal 222.86 us dollars

8 0

3 years ago

Read 2 more answers

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.

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.

3 0

3 years ago

-2.8 repeating as a fraction<br> I really need this in a hurry Thank You!!!

eduard

-2.888888888 = - 26/9

3 0

3 years ago