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Y_Kistochka [10]
3 years ago
11

What is 11 trillion 950 billion 500 million in scientific notation

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0
In order they are

1.1 * 10^12, 9.5 * 10^11 and 5*10^8
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What is value of x? please help
bija089 [108]

Answer:

x = 27

Step-by-step explanation:

To find the value of x, create a proportion. A proportion is two equal ratios for similar figures. The two triangles are similar because they are the same shape but not the same size. Create ratios between corresponding sides of figures.

\frac{x+8}{10}=\frac{2x-5}{14}

To solve for x, cross multiply:

(x+8)(14) = (10)(2x-5)

14x + 112 = 20x -50

        112 = 6x - 50

        162 = 6x

          27 = x

7 0
3 years ago
Can someone plz help me find the surface area of this triangular prism thx
alukav5142 [94]

392.4 in2 should be correct

example of a net...


8 0
3 years ago
According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe
Ann [662]

Using the <em>normal distribution and the central limit theorem</em>, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1 - p)}{n}}, as long as np \geq 10 and n(1 - p) \geq 10.

In this problem:

  • 22% of couples meet online, hence p = 0.22.
  • A sample of 150 couples is taken, hence n = 150.

Item a:

The mean and the standard error are given by:

\mu = p = 0.22

s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{150}} = 0.0338

The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

Item b:

The probability is <u>one subtracted by the p-value of Z when X = 0.25</u>, hence:

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem:

Z = \frac{X - \mu}{s}

Z = \frac{0.25 - 0.22}{0.0338}

Z = 0.89

Z = 0.89 has a p-value of 0.8133.

1 - 0.8133 = 0.1867.

There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

Item c:

The probability is the <u>p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15</u>, hence:

X = 0.2:

Z = \frac{X - \mu}{s}

Z = \frac{0.2 - 0.22}{0.0338}

Z = -0.59

Z = -0.59 has a p-value of 0.2776.

X = 0.15:

Z = \frac{X - \mu}{s}

Z = \frac{0.15 - 0.22}{0.0338}

Z = -2.07

Z = -2.07 has a p-value of 0.0192.

0.2776 - 0.0192 = 0.2584.

There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can check brainly.com/question/24663213

4 0
2 years ago
Three-fourths of p is 18
never [62]
<h3><u>The value of p is 24.</u></h3>

3/4p = 18

Multiply both sides by 4.

3p = 72

Divide both sides by 3.

p = 24


8 0
3 years ago
What’s the answer to this question
Nataly_w [17]
     6 x^{3} + 3x + 5
-    2x^{2} + 6x + 5
-----------------------------------------
      4x - 3 + 0
5 0
3 years ago
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