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Ray Of Light [21]
3 years ago
10

Write the equation of the line, in point-slope form. Identify (x, y) as the point (-2, 2). Use the box provided or the upload

Mathematics
1 answer:
Bingel [31]3 years ago
3 0

Answer:

y = -x + 0

Step-by-step explanation:

well the equation of a line is y = mx + b

m = the slope , b = the y-intercept

m = y2 - y1 / x2 - x1

m = -1

and b is the y-intercept of the line.

finally:

y = -1x + 0

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• Paul has 5 pencils, each x inches in length. When she lines up the pencils end to end, they measure 34.5 inches. Select all th
IgorC [24]

Answer:

B and C

Step-by-step explanation:

Both of these equationa can be used to find the length of each pencil

6 0
3 years ago
March 21, the 80th day of the year, is the spring equinox. find the number of hours of daylight in fairbanks on this day
pentagon [3]
According to a daylight calculator for Fairbanks, AK, sun rises at 07:40 and sun sets at 20:12, giving 12 hours and 32 minutes of daylight.

Note that it is not a mistake that daylight is not exactly 12 hours on that day due to refraction of the sun's rays at sunrise and sunset.

At Fairbanks, near equality to 12 hours occurs on March 16 with sunrise at 07:58 and sunset at 19:57.
3 0
3 years ago
A farmer uses a lot of fertilizer to grow his crops. The farmer's manager thinks fertilizer products from distributor A contain
noname [10]

Answer:

Z_{H0}= 1.56

Step-by-step explanation:

Hello!

You have two independent samples of fertilizers from distributor A and B, and need to test if the average content of nitrogen from fertilizer distributed by A is greater than the average content of nitrogen from fertilizer distributed by B.

Sample 1

X₁: Content of nitrogen of a fertilizer batch distributed by A

n₁= 4 batches

Sample mean X₁[bar]= 23pound/batch

σ₁= 4 poundes/batch

Sample 2

X₂: Content of nitrogen of a fertilizer batch distributed by B

n₂= 4 batches

Sample mean X₂[bar]= 18pound/batch

σ₂= 5 pounds/batch

Both variables have a normal distribution. Now since you have the information on the variables distribution and the values of the population standard deviations, you could use a pooled Z-test.

Your hypothesis are:

H₀: μ₁ ≤ μ₂

H₁: μ₁ > μ₂

The statistic to use is:

Z= <u> X₁[bar] - X₂[bar] - (μ₁ - μ₂) </u>~N(0;1)

       √(δ²₁/n₁ + δ²₂/n₂)

Z_{H0}=<u>   </u>(<u>23 - 18) - 0  </u> = 1.56

         √(16/4 + 25/4)

I hope this helps!

8 0
3 years ago
Which of the following speeds will produce lowest miles per gallon?
Ghella [55]

Answer:

a

Step-by-step explanation:

8 0
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.

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
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