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2 years ago

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Audrey ran 2 1/6 miles yesterday. Today, she ran 1 3/6 miles. Tomorrow, she plans on running 1 4/6 miles less than yesterday and

today combined. How many miles will Audrey run tomorrow?

1 answer:

zubka84 [21]2 years ago

7 0

Answer:

2 miles

Step-by-step explanation:

Add the first 2 terms (2 1/6 and 1 3/6) by adding the whole term and fractional term to get 3 4/6, then subtract 1 4/6 by subtracting 1 from 3 and 4/6 from 4/6 to get 2 0/6, which is just 2.

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Quotien of 6x^4-15x^3-2x^2-10x-4÷3x^2+2

Pachacha [2.7K]

Answer: (6x^4 - 15x^3 - 2x^2 - 10x - 4) ÷ (3x^2 + 2)

= (6x^4 - 15x^3 + 4x^2 - 6x^2 - 10x - 4) ÷ (3x^2 + 2)

= (6x^4 + 4x^2 - 15x^3 - 10x - 6x^2 - 4) ÷ (3x^2 + 2)

= 2x^2(3x^2 + 2) - 5x(3x^2 + 2) - 2(3x^2 + 2) ÷ (3x^2 + 2)

= (3x^2 + 2)(2x^2 - 5x - 2) ÷ (3x^2 + 2)

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2 years ago

Please solve no links i will mark brailest

Lemur [1.5K]

Answer:

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Step-by-step explanation:

Soo... The ratio of blue to green is 4:3 right?

and there are 21 balloons!

4 + 3 is 7! So we divide 21 by 7 and get 3!

Then we multiply 3 by the number of blue balloons(aka 4) and get 12!

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3 years ago

Last winter Armand had 5/6 of a row of stacked logs. At the end of the winter he had 8/15 of the same row left how much wood did

RUDIKE [14]

Answer:

$\frac{3}{10}$

Step-by-step explanation:

First we have to make the fractions even so there denominator have to be equal. So 15 times 2 gets 30 and 6 times 5 gets 30 too. Now multiply the numerator by the corresponding number of 25/30 and 16/30. Subtract to get 9/30 now simplify to get 3/10.

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3 years ago

Read 2 more answers

A selective college would like to have an entering class of 950 students. Because not all students who are offered admission acc

pogonyaev

Answer:

a) The mean is 900 and the standard deviation is 15.

b) 100% probability that at least 800 students accept.

c) 0.05% probability that more than 950 will accept.

d) 94.84% probability that more than 950 will accept

Step-by-step explanation:

We use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

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:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

(a) What are the mean and the standard deviation of the number X of students who accept?

$n = 1200, p = 0.75$ . So

$E(X) = np = 1200*0.75 = 900$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1200*0.75*0.25} = 15$

The mean is 900 and the standard deviation is 15.

(b) Use the Normal approximation to find the probability that at least 800 students accept.

Using continuity corrections, this is $P(X \geq 800 - 0.5) = P(X \geq 799.5)$ , which is 1 subtracted by the pvalue of Z when X = 799.5. So

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

$Z = \frac{799.5 - 900}{15}$

$Z = -6.7$

$Z = -6.7$ has a pvalue of 0.

1 - 0 = 1

100% probability that at least 800 students accept.

(c) The college does not want more than 950 students. What is the probability that more than 950 will accept?

Using continuity corrections, this is $P(X \geq 950 - 0.5) = P(X \geq 949.5)$ , which is 1 subtracted by the pvalue of Z when X = 949.5. So

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

$Z = \frac{949.5 - 900}{15}$

$Z = 3.3$

$Z = 3.3$ has a pvalue of 0.9995

1 - 0.9995 = 0.0005

0.05% probability that more than 950 will accept.

(d) If the college decides to increase the number of admission offers to 1300, what is the probability that more than 950 will accept?

Now n = 1300. So

$E(X) = np = 1300*0.75 = 975$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1200*0.75*0.25} = 15.6$

Same logic as c.

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

$Z = \frac{949.5 - 975}{15.6}$

$Z = -1.63$

$Z = -1.63$ has a pvalue of 0.0516

1 - 0.0516 = 0.9484

94.84% probability that more than 950 will accept

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2 years ago

SOS ANSWERRRRR PLZZZZZ

Karo-lina-s [1.5K]

The answer would be A 20c/d

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2 years ago