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faust18 [17]
3 years ago
6

Pearl uses 9.8 pints of blue paint and white paint to paint her bedroom walls. 2 5 of this amount is blue paint, and the rest is

white paint. How many pints of white paint did she use to paint her bedroom walls?
Mathematics
1 answer:
Ratling [72]3 years ago
7 0

Answer:

She use <u>5.88 pints</u> of white paint.

Step-by-step explanation:

Given:

Pearl uses 9.8 pints of blue paint and white paint to paint her bedroom walls. \frac{2}{5} of this amount is blue paint, and the rest is white paint.

Now, to find the pints of white paint she use.

Pearl uses total pint of paint = 9.8 pints.

The amount of blue paint

= \frac{2}{5}\ of\ 9.8\ pints.

= \frac{2}{5} \times 9.8

= \frac{19.6}{5}

= 3.92\ pints.

Thus, the amount of blue paint = 3.92 pints.

Now, to get the white paint we subtract the amount of blue paint from  the total pint of paint:

9.8-3.92\\\\=5.88\ pints.

Therefore, she use 5.88 pints of white paint.

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Answer:

n^23

Step-by-step explanation:

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6 0
3 years ago
The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a st
Alinara [238K]

Answer:

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

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

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so \mu = 0.5, \sigma = 0.05.

What is the probability that a line width is greater than 0.62 micrometer?

That is P(X > 0.62)

So

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

Z = \frac{0.62 - 0.5}{0.05}

Z = 2.4

Z = 2.4 has a pvalue of 0.99180.

This means that P(X \leq 0.62) = 0.99180.

We also have that

P(X \leq 0.62) + P(X > 0.62) = 1

P(X > 0.62) = 1 - 0.99180 = 0.0082

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

3 0
3 years ago
Solve the equation: k^2+5k+13=0
mr_godi [17]

Step-by-step explanation:

k² + 5k + 13 = 0

Using the quadratic formula which is

x =  \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}  \\

From the question

a = 1 , b = 5 , c = 13

So we have

k =  \frac{ - 5 \pm \sqrt{ {5}^{2} - 4(1)(13) } }{2(1)}  \\  =  \frac{ - 5 \pm \sqrt{25 - 52} }{2}  \\  =  \frac{ - 5 \pm \sqrt{ - 27} }{2}  \:  \:  \:  \:  \:  \:  \\  =  \frac{ - 5  \pm3 \sqrt{3}  \: i}{2}  \:  \:  \:  \:  \:  \:

<u>Separate the solutions</u>

k_1 =  \frac{ - 5 + 3 \sqrt{3} \: i }{2}  \:  \:  \:  \: or \\ k_2 =  \frac{ - 5 - 3 \sqrt{3}  \: i}{2}

The equation has complex roots

<u>Separate the real and imaginary parts</u>

We have the final answer as

k_1 =  -  \frac{5}{2}  +  \frac{3 \sqrt{3} }{2}  \: i \:  \:  \:  \: or \\ k_2 =  -  \frac{5}{2}  -  \frac{3 \sqrt{3} }{2}  \: i

Hope this helps you

8 0
3 years ago
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Wittaler [7]

Answer:

7 means ones because it is the units

Step-by-step explanation:

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6 0
3 years ago
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Mrac [35]

Answer:

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Step-by-step explanation:

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x + 4x = 180

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x = 36

thus exterior angle = 36°

5 0
2 years ago
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