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1 year ago

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The graph shows the stock value for a technology company from 2002 to 2005. From this graph, what is the most likely value of th

e stock for the year 2000? A) $0 B) $10 C) $20 D) $30

1 answer:

deff fn [24]1 year ago

8 0

The answer to your question is B.

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Money Spent on Road Repair A politician wishes to compare the variances of the amount of money spent for road repair in two diff

Mkey [24]

Answer:

(A) Null Hypothesis, $H_0$ : $\sigma_1^{2} =\sigma_2^{2}$ {means that there is no significant difference in the variances of the amounts spent in the two counties}

Alternate Hypothesis, $H_A$ : $\sigma_1^{2} \neq \sigma_2^{2}$ {means that there is a significant difference in the variances of the amounts spent in the two counties}

(B) The F-table gives critical values of 0.359 and 2.781 for (14,17) degrees of freedom for the two-tailed test.

(C) The value of the F-test statistic is 0.611.

(D) We conclude that there is no significant difference in the variances of the amounts spent in the two counties.

Step-by-step explanation:

We are given that a politician wishes to compare the variances of the amount of money spent on road repair in two different counties.

The data are given here below;

County A County B

s1 = $11,596 s2 = $14,837

n1 = 15 n2 = 18

Let $\sigma_1^{2}$ = variances of the amounts spent in County A.

$\sigma_2^{2}$ = variances of the amounts spent in County B.

(A) So, Null Hypothesis, $H_0$ : $\sigma_1^{2} =\sigma_2^{2}$ {means that there is no significant difference in the variances of the amounts spent in the two counties}

Alternate Hypothesis, $H_A$ : $\sigma_1^{2} \neq \sigma_2^{2}$ {means that there is a significant difference in the variances of the amounts spent in the two counties}

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

T.S. = $\frac{s_1^{2} }{s_2^{2} } \times \frac{\sigma_2^{2} }{\sigma_1^{2} }$ ~ $F__n_1_-_1,_ n_2_-_1$

where, $s_1$ = sample standard deviation for County A = $11,596

$s_2$ = sample standard deviation for County B = $14,837

$n_1$ = sample size for County A = 15

$n_2$ = sample size for County B = 18

(B) Now at 0.05 level of significance, the F-table gives critical values of 0.359 and 2.781 for (14,17) degrees of freedom for the two-tailed test.

(C) So, <u><em>the test statistics</em></u> = $\frac{11,596^{2} }{14,837^{2} } \times 1$ ~ $F__1_4,_ 1_7$

= 0.611

The value of the F-test statistic is 0.611.

Now, as we can see that our test statistics lie within the range of critical values of F, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

(D) Therefore, we conclude that there is no significant difference in the variances of the amounts spent in the two counties.

7 0

2 years ago

Triangle PQR with vertices P(3, −6), Q(6, −9), and R(−15, 3) is dilated by a scale factor of 3 to obtain triangle P′Q′R′. Which

vovangra [49]

Answer:

Option C.

Step-by-step explanation:

The vertices of triangle PQR are P(3, −6), Q(6, −9), and R(−15, 3).

It is given that triangle PQR dilated by a scale factor of 3 to obtain triangle P′Q′R′.

We know that a figure and its image after dilation are similar. It means triangle PQR and triangle P′Q′R′

If a figure dilated by factor k about the origin then

$(x,y)\rightarrow (kx,ky)$

PQR dilated by a scale factor of 3, so

$(x,y)\rightarrow (3x,3y)$

Using this rule we get

$P(3,-6)\rightarrow P'(3(3),3(-6))=P'(9,-18)$

$Q(6,-9)\rightarrow Q'(3(6),3(-9))=Q'(18,-27)$

$R(-15,3)\rightarrow R'(3(-15),3(3))=R'(-45,9)$

The vertices of image are P'(9,-18), Q'(18,-27) and R'(-45,9).

Therefore, the correct option is C.

6 0

1 year ago

Caroline bought 20 shares of stock at 10 ½, and after 10 months the value of the stocks was 11 ¼. If Caroline were to sell all h

OleMash [197]

20 shares at 10 1/2 (or 10.5) = 20 * 10.5 = 210
20 shares at 11 1/4 (or 11.25) = 20 * 11.25 = 225

so the profit made would be : 225 - 210 = $ 15 <==

4 0

1 year ago

Ed bought 3 liters of water, 2,750 milliliters of sports drinks, and 2.25 liters of juice. For numbers 4a-4e, select True or Fal

skelet666 [1.2K]

3 liters of water=3,000<span>milliliters of water
</span><span>2.25 liters of juice= 2,250milliliters of juice
</span><span>2,750 milliliters of sports drinks,

</span><span>Ed bought 250 milliliters more water than sports drink it is TRUE
</span>Proof 3,000<span>milliliters -2,750milliliters = 250milliliters

</span><span>Ed bought 1.25 liters more water than juice, it is FALSE
</span>proof 3,000<span>milliliters -2,250milliliters = 750milliliters=0.75l

</span><span>Ed bought 50 milliliters more sports drink than juice. it is FALSE
proof </span><span>2,750 -2,250=500milliliters not 50milliliters

</span><span>Ed bought 0.5 liter more of sports drink than juice. it is TRUE
</span>proof <span>2,750 -2,250=500milliliters =0.5l
</span>
<span>Ed bought 75 milliliters more water than juice. it is FALSE
</span>proof 3,000-2,250= 750milliliters (not 5 milliliters)

6 0

2 years ago

Three ice cream shops have provided quotes for catering Martina’s ice cream party. Each store charges a delivery fee in addition

omeli [17]

First subtract 1.75n-2n-1.25n then you subtract 75 80 and 90 then you divide by n

7 0

2 years ago