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2 years ago

Find the value of x to the nearest tenth.I will mark the brainlest

Mathematics

1 answer:

Feliz [49]2 years ago

8 0

Answer:

The value of x nearest to tenth is, 1.5 units

Step-by-step explanation:

In right angle triangle as shown in figure

by definition of tangent ratio i,e $\tan \theta = \frac{opposite side}{Adjacent side}$

$\tan 37^{\circ} = \frac{x}{2.1}$

$0.7535540501= \frac{x}{2.1}$

$x = 0.7535540501 \times 2.1 = 1.58246351$ units

Therefore, the value of x nearest to tenth is, 1.5 units

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SOLVING QUADRATIC EQUATIONS BY SQUARE ROOT<br> 9p^2-6=642

Aleksandr [31]

9p^2 - 6 = 642; / + 6 ;
9p^2 = 648; / ÷9 ;
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p = + $\sqrt{72} or - \sqrt{72}$ ;
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3 years ago

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3 years ago

2(x+4)=18 help and hurry

daser333 [38]

Answer:

x = 5

Step-by-step explanation:

2x + 8 = 18

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2 years ago

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D = a - b - c solve for b

enot [183]

To isolate b, add a and subtract c to the entire equation:

D - a + c = -b

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2 years ago

A population of values has a normal distribution with μ=204.9μ=204.9 and σ=81.9σ=81.9. You intend to draw a random sample of siz

marin [14]

Answer:

7. μ=204.9 and σ=5.4968

8. μ=75.9 and σ=0.7136

9. p=0.9452

Step-by-step explanation:

7. - Given that the population mean =204.9 and the standard deviation is 81.90 and the sample size n=222.

-The sample mean, $\mu_x$ is calculated as:

$\mu_x=\mu=204.9, \mu_x=sample \ mean$

-The standard deviation, $\sigma_x$ is calculated as:

$\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{81.9}{\sqrt{222}}\\\\=5.4968$

8. For a random variable X.

-Given a X's population mean is 75.9, standard deviation is 9.6 and a sample size of 181

-#The sample mean, $\mu_x$ is calculated as:

$\mu_x=\mu\\\\=75.9$

#The sample standard deviation is calculated as follows:

$\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{9.6}{\sqrt{181}}\\\\=0.7136$

9. Given the population mean, μ=135.7 and σ=88 and n=59

#We calculate the sample mean;

$\mu_x=\mu=135.7$

#Sample standard deviation:

$\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{88}{\sqrt{59}}\\\\=11.4566$

#The sample size, n=59 is at least 30, so we apply Central Limit Theorem:

$P(\bar X>117.4)=P(Z>\frac{117.4-\mu_{\bar x}}{\sigma_x})\\\\=P(Z>\frac{117.4-135.7}{11.4566})\\\\=P(Z>-1.5973)\\\\=1-0.05480 \\\\=0.9452$

Hence, the probability of a random sample's mean being greater than 117.4 is 0.9452

7 0

3 years ago