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Marina86 [1]
3 years ago
15

a few years ago, melky bought 100 shares of a cologne company's stock for $16.77 per share. Last month she sold all of the share

s for $11.88 per share. What was her loss?
Mathematics
2 answers:
garik1379 [7]3 years ago
3 0
$489
$16.77*100
-
$11.88*100
=$489
Taya2010 [7]3 years ago
3 0

Answer:

$489.

Step-by-step explanation:

We have been given that a few years ago, Melky bought 100 shares of a cologne company's stock for $16.77 per share. Last month she sold all of the shares for $11.88 per share.

First of all, we will find the difference of purchase price of per share and selling price of per share.

\text{Loss per share}=\$16.77-\$11.88

\text{Loss per share}=\$4.89

\textMelky's loss}=\$4.89\times 100

\textMelky's loss}=\$489

Therefore, Melky lost $489 in total.

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Which is greater? 4 tenths and 2 hundredths or 2 tenths and 4 hundredths
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.42 is greater than .24

.42  > .24

Hope this helps.

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Need help with trig question
Solnce55 [7]

Answer:

  0 +256i

Step-by-step explanation:

According to Euler's formula, ...

  (4 cis π/8)^4 = (4^4) cis (4×π/8) = 256 cis π/2 = 0 +256i

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"cis" is an abbreviation sometimes used for "cosine + i×sine". It simplifies writing the expression. Engineers sometimes simplify it further, writing 4∠(π/8) for the expression in this problem statement.

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7 0
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Read 2 more answers
Some researchers have conjectured that stem-pitting disease in peach-tree seedlings might be controlled with weed and soil treat
stiks02 [169]

Answer:

Part A

b. 14.6 ± 7.38

Part B

b. 3.43

Part C

a. P-value < 0.01

Part D

b. There is sufficient evidence to reject the null hypothesis

Step-by-step explanation:

Part A

The given data are;

The number of seedlings in the field = 20

The number of seedlings selected to receive herbicide A = 10

The number of seedlings selected to receive herbicide B = 10

The height in centimeters of seedlings treated with Herbicide A, \overline x _1 = 94.5 cm

The standard deviation, s₁ = 10 cm

The height in centimeters of seedlings treated with Herbicide B, \overline x _2 = 109.1 cm

The standard deviation, s₂ = 9 cm

The 90% confidence interval for μ₂ - μ₁, is given as follows;

\left (\bar{x}_{2}- \bar{x}_{1}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

The critical-t at 95% and n₁ + n₂ - 2 degrees of freedom is given as follows;

The degrees of freedom, df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18

α = 100% - 90% = 10%

∴ For two tailed test, we have, α/2 = 10%/2 = 5% = 0.05

t_{(0.025, \, 18)} = 1.734

C.I. = \left (109.1- 94.5 \right )\pm 1.734 \times \sqrt{\dfrac{10^{2}}{10}+\dfrac{9^{2}}{10}}

C.I. ≈ 14.6 ± 7.37714603353

The 90% C.I. ≈ 14.6 ± 7.38

b. 14.6 ± 7.38

Part B

With the hypotheses are given as follows;

H₀; μ₂ - μ₁ = 0

Hₐ; μ₂ - μ₁ ≠ 0

The two sample t-statistic is given as follows;

t=\dfrac{(\bar{x}_{2}-\bar{x}_{1})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}+\dfrac{s _{2}^{2}}{n_{2}}}}

t-statistic=\dfrac{(109.1-94.5)}{\sqrt{\dfrac{10^{2} }{10}+\dfrac{9^{2}}{10}}} \approx 3.43173361147

The two sample t-statistic ≈ 3.43

b. 3.43

Part C

From the t-table, the p-value, we have, the p-value < 0.01

a. P-value < 0.01

Part D

Given that a significance level of 0.05 level is used and the p-value of 0.01 is less than the significance level, there is enough statistical evidence to reject the null hypothesis

b. There is sufficient evidence to reject the null hypothesis.

4 0
3 years ago
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bekas [8.4K]

Answer:

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Step-by-step explanation:

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=> 6m(n)^3 + 3m(n)^3 + 15m(n)^2 - m(n)^2

=> 9m(n)^3 +14m(n)^2

4 0
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