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3 years ago

Solve the following equation. Then place the correct number in the box provided. 3x + 1 + 5x = 7 +15 + 7x find X

Mathematics

1 answer:

KengaRu [80]3 years ago

7 0

3x + 1 + 5x = 7 + 15 + 7x
8x + 1 = 22 + 7x
x = 21

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Mary, Chau, and David have a total of $87 i their wallets. Marry has 9$ more than Chau. David has two times what Mary has. How m

Pavel [41]

Answer:

Mary = 24

Chau = 15

David = 48

Step-by-step explanation:

The formula is

Mary + Chau + David = 87

And we know that

Chau = Mary - 9

David = Mary * 2

So when we fill this in

Mary + Mary - 9 + Mary * 2 = 87

4Mary - 9 = 87

4Mary = 96

Mary = 24

Chau = Mary - 9 = 15

David = Mary * 2 = 48

7 0

2 years ago

9. In a 25-kilometer triathlon, competitors sim 2 kilometers, run 5 kilometers, and bike the rest

postnew [5]

Answer:

They bike 18 kilometers

Step-by-step explanation:

25-2+5=18

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3 years ago

What is the solution?<br><img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%200%7D%20%5Cfrac%7Bx%20%2B%202%7D%7B%20x%20-2%7D" id

zhenek [66]

Answer:

$- 1$

Step-by-step explanation:

$\lim_{x\to 0} \frac{x + 2}{ x -2}$

$= \frac{0 + 2}{ 0 -2}$

$= \frac{2}{-2}$

$= - 1$

5 0

2 years ago

Read 2 more answers

Austin wants to burn a total of 700 calories. So far, he has burned 581.4 calories. How many more calories must Austin burn?

Naddik [55]

Answer:

118.6

Step-by-step explanation:

Total calories to burn = 700

Total calories burnt = 581.4

Remaining Calories = 700 - 581.4 = 118.6

8 0

2 years ago

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A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test. What is

Artemon [7]

Answer:

The 95% confidence interval for the mean score, , of all students taking the test is

$28.37< L\ 30.63$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 59$

The mean score is $\= x = 29.5$

The standard deviation $\sigma = 5.2$

Generally the standard deviation of mean is mathematically represented as

$\sigma _{\= x} = \frac{\sigma }{\sqrt{n} }$

substituting values

$\sigma _{\= x} = \frac{5.2 }{\sqrt{59} }$

$\sigma _{\= x} = 0.677$

The degree of freedom is mathematically represented as

$df = n - 1$

substituting values

$df = 59 -1$

$df = 58$

Given that the confidence interval is 95% then the level of significance is mathematically represented as

$\alpha = 100 -95$

$\alpha =$ 5%

$\alpha = 0.05$

Now the critical value at this significance level and degree of freedom is

$t_{df , \alpha } = t_{58, 0.05 } = 1.672$

Obtained from the critical value table

So the the 95% confidence interval for the mean score, , of all students taking the test is mathematically represented as

$\= x - t*(\sigma_{\= x}) < L\ \= x + t*(\sigma_{\= x})$

substituting value

$(29.5 - 1.672* 0.677) < L\ (29.5 + 1.672* 0.677)$

$28.37< L\ 30.63$

4 0

3 years ago