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zzz [600]
3 years ago
10

Max brought $20 to the arcade. Each arcade game cost $1.25 to play. He used $6.50 to buy snacks. Which of the following inequali

ty equations below shows the greatest number of arcade games Max can play with the money he brought?
A. 1.25g - 6.50 > 20
B. 1.25g + 6.50 > 20
C. 1.25g + 6.50 < 20
D. 1.25 - 6.50g < 20
Mathematics
2 answers:
Ronch [10]3 years ago
7 0
Is B actually the right one?
Greeley [361]3 years ago
4 0

Answer:

B

Step-by-step explanation:

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Graph this function: y–3= – (x+10) <br> Click to select points on the graph.
Nikolay [14]

Answer:

See below

Step-by-step explanation:

We can simply use a graphing calculator to graph the function

\boxed{y-3 = -(x+10)}

See in the attached file!

7 0
2 years ago
Several paint mixtures are made by mixing blue paint and white paint. Paint Colors Blue Paint White Paint Mixture A 5 cups 12 cu
Nesterboy [21]

Answer:

The mixture C is the correct option

Step-by-step explanation:

According to the given scenario, the calculation is as follows:

For Mixture A

Blue Paint - 5 cups

White Paint - 12 cups

The ratio between them is 5:12

For Mixture B

Blue Paint - 6 cups

White Paint - 6 cups

 The ratio between them is 6:6 = 12:12

It came by multiply the numerator and denominator by 12

For Mixture C

Blue Paint - 4 cups

White Paint - 12 cups

 The ratio between them is 4:12

For Mixture D

Blue Paint - 5 cups

White Paint - 6 cups

 The ratio between them is 5:6 = 10:12

It came by multiply the numerator and denominator by 12

As it can be seen that in all four mixtures the denominator is the same so for calculating the lowest ratio we have to see the small value in the numerator

As it can be seen that there is a small value of 4

hence, the mixture C is the correct option

5 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
Find the median.<br> 2, 20, 19, 15, 11,9,7,1<br> OA. 10<br> OB.<br> 10.5<br> c. 19<br> D. 20
pickupchik [31]

Answer:

A. 10

Step-by-step explanation:

First put the numbers in order from least to greatest then find the two numbers in the center of that since there is an even number of numbers there will be two the add and divide by two and you get 10.

1,2,7,9,11,15,19,20

9+11=20

20/2=10

3 0
3 years ago
Read 2 more answers
Consider the following data:
-BARSIC- [3]

Answer: x =6

Step-by-step explanation:

8 0
2 years ago
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