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3 years ago

15

Together, Tom and max have 72 football cards. Tom has 2 more than 4 times as many cards as max has. How many football cards does

Tom have?

1 answer:

frosja888 [35]3 years ago

4 0

Tom has 58 football cards

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Lillianne made 6 out of every 10 baskets she attempted during basketball practice. If she attempted to make 25 baskets, how many

patriot [66]

Answer:

15.

Step-by-step explanation:

(6/ 10) * 25

= 0.6 * 25

= 15.

3 0

3 years ago

What is the image of (-8,6) after a translation over the line y=-x

stira [4]

Answer:

(-6, 8)

Step-by-step explanation:

(x, y) --> (-y, -x)

(-8, 6) --> (-6, 8)

Therefore, (-6, 8)

I hope this helped and have a good rest of your day!

8 0

3 years ago

Cole is studying employment trends.

natali 33 [55]

Answer:

225

Step-by-step explanation:

So we know that 20% of the social workers he surveyed was 45 people. The easiest way to find all 100% of the social workers is to do 45 x 5 because 20% x 5 = 100%. So 45 x 5 = 225.

Hope this helped! :)

3 0

3 years ago

12x>9 (2x-3)-15 solve for x and show steps please.

Verdich [7]

12x > 9(2x-3)-15

distribute the 9

12x > 18x-27 -15

combine like terms on the right side

12x > 18x-42

subtract 18x from both sides

-6x > -42

divide both sides by -6

x = -42 / -6 = 7

x >7

4 0

3 years ago

Read 2 more answers

Suppose the horses in a large stable have a mean weight of 975lbs, and a standard deviation of 52lbs. What is the probability th

Lubov Fominskaja [6]

Answer:

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 975, \sigma = 52, n = 31, s = \frac{52}{\sqrt{31}} = 9.34$

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable?

pvalue of Z when X = 975 + 15 = 990 subtracted by the pvalue of Z when X = 975 - 15 = 960. So

X = 990

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

By the Central Limit Theorem

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

$Z = \frac{990 - 975}{9.34}$

$Z = 1.61$

$Z = 1.61$ has a pvalue of 0.9463

X = 960

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

$Z = \frac{960 - 975}{9.34}$

$Z = -1.61$

$Z = -1.61$ has a pvalue of 0.0537

0.9463 - 0.0537 = 0.8926

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

7 0

3 years ago