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elixir [45]
3 years ago
11

nicole worked 13 hours each day for 9 days. then the next day she worked 6 hours. how many hours did nicole work in all???

Mathematics
2 answers:
alex41 [277]3 years ago
4 0

13(9)+6=117+6=123

She worked for 123h.

cestrela7 [59]3 years ago
4 0

Nicole worked 123 hours.


Now let me show you how i got that!


First you want to set up your equation.

13 times 9 plus 6= x

13 times 9= 117

117+6

117+6=123

x=123

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Mandarinka [93]

Answer:

<h2>It is 1</h2>

Step-by-step explanation:

{7}^{1}  = 7

I hope that is useful for you :)

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3 years ago
If the length of each side of a cuboid decreases by 20%, find the percentage decrease in its volume.
just olya [345]

Answer:

Step-by-step explanation:

(1+25 /100) (1-20/100) (1-50/100)  <1

5/4 x 4/5 x 1/2 <1

Decrease in volume (in percent)

(1+25 /100) (1-20/100) (1-50/100)  x 100

=48.8%

6 0
3 years ago
Please help me
elena55 [62]

Answer:

a)

<u>The equation would be:</u>

  • 3x + 2 = 7*2

b)

Step 1. <u>Subtract 2 from both sides:</u>

  • 3x = 6*2

Step 2. <u>Divide both sides by 3:</u>

  • 3x/3 = 6*2/3
  • x = 2*2

c)

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6 0
3 years ago
Read 2 more answers
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

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3 years ago
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