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Alex
3 years ago
12

Each side of a square is increasing at a rate of 6 cm/s. At what rate is the area of the square increasing when the area of the

square is 16cm^2?
Mathematics
1 answer:
siniylev [52]3 years ago
7 0

<u>Solution:</u> 48cm/s^2

<u>Working:</u>

A = s^2

Derivate s^2 (as it is area formula) which gives

dA/dt = 2s dL/dt

dL/dt = 6cm/s

Hence,

2(6) = 12

Side: 4cm

Hence,

dA/dt = (4)(12) = 48cm/s^2

<em>Feel free to mark this as brainliest :D</em>

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Answer:

here my answer

Step-by-step explanation:

5+1=6

7-1=6

3x2=6

hope it helps :)

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What is the image of (8,7) after a reflection over the line y=-x?
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Answer:

(-7,-8)

Step-by-step explanation:

The original point is (8,7)

To reflect it, just switch their places and change their signs.

So, (-7,-8)

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2. Geoff planted dahlias in his garden. Dahlias have bulbs that
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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
Write in standard form <br><br><br>5.13 x 10₇<br><br>*please help me*
mafiozo [28]

In the standard form, the number 7 here should be an exponent not a sub text.

Answer: 5.13 x 10^(7).

Done!

4 0
3 years ago
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