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cricket20 [7]
2 years ago
9

Consider parallelogram ABCD with diagonal BD. What values of x and y prove that the parallelogram is a square if CD=20, AD=8x−4,

and ∠CBD=4y−3?
a. X = 2, y = 12

b. X = 2, y = 23. 5

c. X = 3, y = 12

d. X = 3, y = 23. 5
Mathematics
1 answer:
Radda [10]2 years ago
7 0

Answer: C

Step-by-step explanation:

For the parallelogram to be a square, all sides must have equal measure and all angles must be right.

This means that CD=AD -> 20=8x-4 -> 24=8x -> x=3

And 4y-3=45 -> 4y=48 -> y=12

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The Kellogg Company periodically compares sales across departments. In one particular instance, they would like to determine if
alex41 [277]

Answer:

H₀: μs = μf

H₁: μs ≠ μf

Step-by-step explanation:

Hello!

If the company wants to compare the sales of both departments "snacks" and "frozen foods" to see if there is any difference, the best is to compare the average number of sales of them.

Then the parameters of interest will be:

μs= average number of sales of the "snacks" department.

μf= average number of sales of the "frozen food" department.

The objective is to test if there is any difference between both departments, then the hypothesis should be two-tailed:

H₀: μs = μf

H₁: μs ≠ μf

I hope it helps!

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3 years ago
Hurry if correct I will give brainly ​
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D

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6 0
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The price of a technology stock has dropped to $9.64 today. Yesterday's price was $9.75. Find the percentage decrease. Round you
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Step-by-step explanation:

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7 0
2 years ago
The pesticide diazinon is in common use to treat infestations of the German cockroach, Blattella germanica. A study investigated
cluponka [151]

Answer:

We conclude that there is no difference in the proportion of cockroaches that died on each surface.

Step-by-step explanation:

We are given that a study investigated the persistence of this pesticide on various types of surfaces.

After 14 days, they randomly assigned 72 cockroaches to two groups of 36, placed one group on each surface, and recorded the number that died within 48 hours. On the glass, 18 cockroaches died, while on plasterboard, 25 died.

<em>Let </em>p_1<em> = proportion of cockroaches that died on glass surface.</em>

<em />p_2<em> = proportion of cockroaches that died on plasterboard surface.</em>

So, Null Hypothesis, H_0 : p_1-p_2 = 0      {means that there is no difference in the proportion of cockroaches that died on each surface}

Alternate Hypothesis, H_A : p_1-p_2\neq 0      {means that there is a significant difference in the proportion of cockroaches that died on each surface}

The test statistics that would be used here <u>Two-sample z proportion</u> <u>statistics</u>;

                       T.S. =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}  } }  ~ N(0,1)

where, \hat p_1 = sample proportion of cockroaches that died on glass surface = \frac{18}{36} = 0.50

\hat p_2 = sample proportion of cockroaches that died on plasterboard surface = \frac{25}{36} = 0.694

n_1 = sample of cockroaches on glass surface = 36

n_2 = sample of cockroaches on plasterboard surface = 36

So, <u><em>test statistics</em></u>  =  \frac{(0.50-0.694)-(0)}{\sqrt{\frac{0.50(1-0.50)}{36}+\frac{0.694(1-0.694)}{36}  } }

                               =  -1.712

The value of z test statistics is -1.712.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that there is no difference in the proportion of cockroaches that died on each surface.

8 0
4 years ago
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