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LenaWriter [7]
3 years ago
11

A certain television is advertised as a 41-inch TV (the diagonal length). If the width of the TV is 40 inches, how many inches t

all is the TV?
Mathematics
2 answers:
Bumek [7]3 years ago
6 0

Answer:

9 inches tall

Step-by-step explanation:

Use the Pythagorean theorem: you know the hypotenuse (41) and the leg (40)

a^{2} +40^{2} =41^{2}

Solve for a:

Simplify exponents

a^2+1600=1681

Subtract 1600 from both sides

a^2+1600-1600=1681-1600\\a^2=81

Find the square root

\sqrt{a^2}=\sqrt{81\\}  \\a=9

KengaRu [80]3 years ago
5 0

Answer:

9 inches

Step-by-step explanation:

A^2+b^2=c^2 so

40^2+b^2=41^2

1600+b^2=1681

so b^2=81

so b=9

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4 years ago
2a+3/2 - a-2/3 = a-1/4
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Answer:

No solutions

Step-by-step explanation:

2a+\dfrac{3}{2}-a-\dfrac{2}{3}=a-\dfrac{1}{4}\\a+\dfrac{5}{9}=a-\dfrac{1}{4}\\\dfrac{5}{9}=\dfrac{1}{4}

This equation has no solutions. Hope this helps!

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3 years ago
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2.A production process manufactures items with weights that are normally distributed with mean 10 pounds and standard deviation
Vesna [10]

Answer:

Step-by-step explanation:

Given that:

population mean = 10

standard deviation = 0.1

sample mean = 9.8 < x > 10.2

The z score can be computed as:

z = \dfrac{\bar x - \mu}{\sigma}

if x > 10.2

z = \dfrac{10.2- 10}{0.1}

z = \dfrac{0.2}{0.1}

z = 2

If x < 9.8

z = \dfrac{9.8- 10}{0.1}

z = \dfrac{-0.2}{0.1}

z = -2

The p-value = P (z ≤ 2) + P (z ≥ 2)

The p-value = P (z ≤ 2) + ( 1 -  P (z ≥ 2)

p-value = 0.022750 +(1 -   0.97725)

p-value = 0.022750 +  0.022750

p-value = 0.0455

Therefore; the probability of defectives  = 4.55%

the probability of acceptable = 1 - the probability of defectives

the probability of acceptable = 1 - 0.0455

the probability of acceptable = 0.9545

the probability of acceptable = 95.45%

4.55% are defective or 95.45% is acceptable.

sampling distribution of proportions:

sample size n=1000

p = 0.0455

The z - score for this distribution at most 5% of the items is;

z = \dfrac{0.05 - 0.0455}{\sqrt{\dfrac{0.0455\times 0.9545}{1000}}}

z = \dfrac{0.0045}{\sqrt{\dfrac{0.04342975}{1000}}}

z = \dfrac{0.0045}{\sqrt{4.342975 \times 10^{-5}}}

z = 0.6828

The p-value = P(z ≤ 0.6828)

From the z tables

p-value = 0.7526

Thus, the probability that at most 5% of the items in a given batch will be defective = 0.7526

The z - score for this distribution for at least 85% of the items is;

z = \dfrac{0.85 - 0.9545}{\sqrt{\dfrac{0.0455\times 0.9545}{1000}}}

z = \dfrac{-0.1045}{\sqrt{\dfrac{0.04342975}{1000}}}

z = −15.86

p-value = P(z ≥  -15.86)

p-value = 1 - P(z <  -15.86)

p-value = 1 - 0

p-value = 1

Thus, the probability that at least 85% of these items in a given batch will be acceptable = 1

6 0
3 years ago
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