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3 years ago

10 points for answering! What is the answer to 13, 14, and 15?

Mathematics

1 answer:

ki77a [65]3 years ago

4 0

Step-by-step explanation:

$\implies\sf{ {x}^{2} = 144 } \\ \\ \implies\sf{ x = \pm \sqrt{144} } \\ \\ \implies\sf{ x = \pm 12 }$

$\implies\sf{ {x}^{2} = \dfrac{25}{289} } \\ \\ \implies\sf{ x = \pm \sqrt{ \frac{25}{289} } } \\ \\ \implies\sf{ x = \pm \frac{5}{17} }$

$\implies\sf{ {x}^{3} = 216 } \\ \\ \implies\sf{ x = \sqrt[3]{216} } \\ \\ \implies\sf{ x = 6 }$

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Nicki dashed up the shore and into the forest, not stopping

romanna [79]

Answer:

Step-by-step explanation:

the answer is not revealed

5 0

3 years ago

Ian picked 5 times as many berries as Michelle.Joe picked 7 times as many berries Michelle.Together they picked 208 berries.How

Rasek [7]

Michelle: x
Ian: 5x
Joe: 7x
Michelle + Ian + Joe = 208 berries
x + 5x + 7x = 208
Combine like terms: 13x = 208
Divide both sides by 13: x = 16
Joe = 7x = 7*16=112
Joe picked 112 berries

8 0

3 years ago

Read 2 more answers

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

Mamont248 [21]

Answer:

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 50, \sigma = 6$

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

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

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

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

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

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

$Z = \frac{67 - 50}{6}$

$Z = 2.83$

$Z = 2.83$ has a pvalue 0.9977

X = 60

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

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

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

$Z = \frac{76 - 50}{6}$

$Z = 4.33$

$Z = 4.33$ has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

e. To be among the 10% fastest snails, a snail must finish in at most hours.

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

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

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

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

$-1.28 = \frac{X - 50}{6}$

$X - 50 = -1.28*6$

$X = 42.32$

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

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

$1.28 = \frac{X - 50}{6}$

$X - 50 = 1.28*6$

$X = 57.68$

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

5 0

3 years ago

Which values of a and b in the exponential function y = a times b Superscript x would result in the following graph? On a coordi

galina1969 [7]

Answer:

$a =-3$ $b =2$

Step-by-step explanation:

Given

$y = ab^x$

$(x_1,y_1) = (0,-3)$

Required

Find a and b

Substitute $(x_1,y_1) = (0,-3)$ for x and y in $y = ab^x$

$-3 = ab^0$

$-3 = a * 1$

$-3 = a$

Rewrite as:

$a =-3$

The above implies that (a) is correct;

Hence:

$a =-3$ $b =2$

So, the equation is:

$y = -3(2)^x$

6 0

2 years ago

Read 2 more answers

Sabendo que x2 + y2=74 e que xy=35, calcule valor de (x - y2)2

Ann [662]

To solve this problem you must apply the proccedure shown below:

1. You have the following information given in the problem above:

- x²+y²<span>=74
xy=35

2. Therefore, you have that (x-y)</span>² is:

=(x-y)²
=x²-2xy+y²
=(x²+y²)-2xy

3. When you substitute the values given in the problem, you obtain:

=74-2(35)
=74-70
=4

4. Therefore, as you can see, the correct answer is: 4

6 0

3 years ago