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2 years ago

Given 4x − 8y = 8:

1 answer:

3 0

4x - 8y = 8
8y = 4x - 8
y = 1/2 x - 1

slope: 1/2
y-intercept:(0, -1)

Line perpendicular- negative reciprocal slope

y-(-1) = -2 (x-0) Hahaha I don't know what is the point slope form though hahaha not sure with this.

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You received a gift in the mail. The length of the box is 12 inches, the width is 8

Softa [21]

Answer: 296 or 592

Step-by-step explanation:

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2 years ago

A sphere has a radius of 6 meters. A second sphere has a radius of 3 meters. What is the difference of the volumes of the sphere

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The volume of the first figure: 904.78 m^3
the volume of the second figure: 113.1 m^3
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3 years ago

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What is 3*5/2 A.5/6 B.15/2 C.8/2 D.15/6

alina1380 [7]

Answer:

The correct answer to this question is B

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2 years ago

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Two vertical poles of length 10 and 12 feet, respectively, stand 17 feet apart. A cable reaches from the top of one pole to the

Marat540 [252]

Answer:

$L=\sqrt{x^{2}+100}+\sqrt{x^{2}-34x+433} ft$

Step-by-step explanation:

let length of calble = L

distance from 10-foot pole = x ft

Moving from the top of the 10-foot pole to point x, we have

From the diagram that

$x^{2} +10^{2}=L_{1} ^{2} \\x^{2} + 100 = L_{1} ^{2}$

Take the square root of both sides

$\sqrt{x^{2}+100}=\sqrt{L_{1} ^{2}}\\ \sqrt{x^{2}+100}= L_{1}$

Moving from point x to the top of the 12-foot pole, we have

From the diagram that

$(17-x)^{2}+12^{2}=L_{2} ^{2} \\289-34x+x^{2} +144=L_{2} ^{2}\\x^{2} -34x+433=L_{2} ^{2}$

Take the square root of both sides

$\sqrt{x^{2}-34x+433}=\sqrt{L_{2} ^{2}} \\\sqrt{x^{2}-34x+433}=L_{2}$

Amount of cable used, L = L1 + L2

$L=\sqrt{x^{2}+100}+\sqrt{x^{2}-34x+433}$

3 0

3 years ago

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73

faust18 [17]

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62 [from the standard t table]

Therefore,

CI = $(Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})$

Thus,

Lower limit of CI = $(Mean - \frac{t\times\sigma}{\sqrt{n}})$

Lower limit of CI = $(98.3 - \frac{2.62\times0.73}{\sqrt{103}})$

Lower limit of CI = 98.11

and,

Upper limit of CI = $(Mean + \frac{t\times\sigma}{\sqrt{n}})$

Upper limit of CI = $(98.3 + \frac{2.62\times0.73}{\sqrt{103}})$

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and the mean temperature could very possibly be 98.6°F

7 0

2 years ago