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Olin [163]
3 years ago
6

a shark can swim at an average speed of 25 miles per hour. at this rate, how far can a shark swim in 2.4 hours? use r = d/t

Mathematics
1 answer:
nirvana33 [79]3 years ago
3 0
2.4 *60mins=144
25*60mins=1500

r=d/t
r= 1500/144
r=10.4

or

25/2.4= 10.4
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Of the total amount of available water on the earth, what percent is usable?
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Expand the following:<br> a) x(x + 2)<br> b) x(2x - 5)<br> c) 2x(3x + 4)<br> d) 6x(x - 2)
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Answer:

<h2>a. \:  {x}^{2}  + 2</h2>

<h2> </h2><h2>b. \: 2 {x}^{2}  - 5x</h2><h2> </h2><h2>c. \: 6 {x}^{2}  + 8x</h2><h3> </h3><h3>d. \: 6 {x}^{2}  - 12x</h3>

solution,

a. \: x(x + 2) \\  \:  \:  = x \times x + 2 \times x \\  \:  \:  =  {x}^{2}  + 2x

b . \: x(2x - 5) \\  \:  = x \times 2x - x \times 5 \\  \:  \:  = 2 {x}^{2}  - 5x

c. \: 2x(3x + 4) \\  \:  = 2x \times 3x + 2x \times 4 \\  \:  \:  = 6 {x}^{2}  + 8x

d. \: 6x(x - 2) \\  \:  \:  = 6x \times x  - 6x \times 2 \\  \:  \:  = 6 {x}^{2}  - 12x

<h2> </h2><h2 />

Hope this helps...

Good luck on your assignment..

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3 years ago
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3 years ago
Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two​ bakeries' specialty pi
DIA [1.3K]

Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (<u>49</u>) < μ₁ - μ₂ < (<u>289)</u>

Step-by-step explanation:

The given data are;

Bakery A

\overline x_1<em> </em>= 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

\overline x_2<em> </em>= 1,711 cal

s₂ = 192 cal

n₂ = 10

\left (\bar{x}_1-\bar{x}_{2}  \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2}  \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}

df = n₁ + n₂ - 2

∴ df = 10 + 18 - 2 = 26

From the t-table, we have, for two tails, t_c = 1.706

\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}

\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469

\hat \sigma ≈ 178

Therefore, we get;

\left (1,880-1,711  \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711  \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}

Which gives;

169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}

Therefore, by rounding to the nearest integer, we have;

The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289

4 0
3 years ago
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