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1 year ago

Please help with math problem?

Mathematics

1 answer:

Hoochie [10]1 year ago

5 0

Answer:

KM=41.6

Step-by-step explanation:

$\frac{HJ}{IJ} =\frac{KM}{ML} \\\frac{8}{6} =\frac{KM}{31.2} \\$

Cross Multiply and Solve for KM.

$6*KM=8*31.2\\KM=\frac{8*31.2}{6} \\KM=41.6$

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Dangers drinking alcohol in young people today and causes of it

Minchanka [31]

Answer:

School problems, such as higher rates of absences or lower grades.

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hope this helps

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Step-by-step explanation:

4 0

2 years ago

Pediatricians work an average of 48 h per week. The standard deviation is 12 hours. What percentage of pediatricians work more t

cricket20 [7]

Answer: 2.28 %

Step-by-step explanation:

Given : Mean : $\mu=\ 48$

Standard deviation : $\sigma= 12$

The formula to calculate the z-score :-

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

For x = 72

$z=\dfrac{72-48}{12}=2$

The p-value = $P(2$

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In percent, $0.0228\times100=2.28\%$

Hence, the percentage of pediatricians work more than 72 h per week = 2.28 %

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2 years ago

Sketch the line passing through the pair of points and identify the line as having a positive slope, a negative slope, a zero sl

scoundrel [369]

Answer:

line will be horizontal running through (0,2)

Step-by-step explanation:

6,2
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1 year ago

219.44 in word form

Naily [24]

Answer:

two hundred nineteen and 44

two hundred nineteen point 44

Step-by-step explanation:

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6 0

2 years ago

Read 2 more answers

Find the general indefinite integral. (Use C for the constant of integration. Remember to use absolute values where appropriate.

Maru [420]

Answer:

$9\text{ln}|x|+2\sqrt{x}+x+C$

Step-by-step explanation:

We have been an integral $\int \frac{9+\sqrt{x}+x}{x}dx$ . We are asked to find the general solution for the given indefinite integral.

We can rewrite our given integral as:

$\int \frac{9}{x}+\frac{\sqrt{x}}{x}+\frac{x}{x}dx$

$\int \frac{9}{x}+\frac{1}{\sqrt{x}}+1dx$

Now, we will apply the sum rule of integrals as:

$\int \frac{9}{x}dx+\int \frac{1}{\sqrt{x}}dx+\int 1dx$

$9\int \frac{1}{x}dx+\int x^{-\frac{1}{2}}dx+\int 1dx$

Using common integral $\int \frac{1}{x}dx=\text{ln}|x|$ , we will get:

$9\text{ln}|x|+\int x^{-\frac{1}{2}}dx+\int 1dx$

Now, we will use power rule of integrals as:

$9\text{ln}|x|+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\int 1dx$

$9\text{ln}|x|+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+\int 1dx$

$9\text{ln}|x|+2x^{\frac{1}{2}}+\int 1dx$

$9\text{ln}|x|+2\sqrt{x}+\int 1dx$

We know that integral of a constant is equal to constant times x, so integral of 1 would be x.

$9\text{ln}|x|+2\sqrt{x}+x+C$

Therefore, our required integral would be $9\text{ln}|x|+2\sqrt{x}+x+C$ .

4 0

2 years ago