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4 years ago

Please need help to answer the question in 5 minutes?

Mathematics

1 answer:

Sergio039 [100]4 years ago

3 0

Answer:

22 7/18 pounds

Step-by-step explanation:

A: 15 1/2 pounds

B: 15 1/2 x 1/3 = 5 1/6

C: 5 1/6 x 1/3 = 1 13/18

A+B+C = 15 1/2 + 5 1/6 + 1 13/18 = 403/18 = 22 7/18

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The sum of two numbers is 90. The larger number is 14 more than 3 times the smaller number. Find the numbers

e-lub [12.9K]

Answer:

The numbers are 19 and 71.

Step-by-step explanation:

Let x and y represent the two numbers. Then x + y = 90.

Also, if we assume that the larger number is y, then y = 3x + 14.

We must solve this system of linear equations. Since we already have the result of solution for y (y = 3x + 14), let's use the substitution method to find the solution:

x + y = 90 becomes x + 3x + 14 = 90, or:

4x = 76. Thus, x = 76/4, or 19. If x = 19, then y = 3(19) + 14, or y = 71.

The numbers are 19 and 71.

Note that these add up to 90, as they must, and that 71 is 14 more than 3 times 19.

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3 years ago

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Olin [163]

Answer:

-5n-6n< 8-8n-n

-11n < -9n + 8

-2n < 8

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3 years ago

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What is the equation of the line whose slope is 20 and a y-intercept of 5

aleksandr82 [10.1K]

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3 years ago

A sample of 900 college freshmen were randomly selected for a national survey. Among the survey participants, 372 students were

Rufina [12.5K]

Answer:

a) $ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423$

b) $0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371$

Step-by-step explanation:

1) Data given and notation

n=900 represent the random sample taken

X=372 represent the students were pursuing liberal arts degrees

$\hat p=\frac{372}{900}=0.413$ estimated proportion of students were pursuing liberal arts degrees

$\alpha=0.01$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

p= population proportion of students were pursuing liberal arts degrees

2) Solution to the problem

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.