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Andru [333]
3 years ago
8

Admission to carnival is $5 and each game at the carnival cost $0.85 you have $15 dollars to spend on admission and games. What

is the maximum number of games you can play?

Mathematics
1 answer:
Serhud [2]3 years ago
4 0
First you need an equation
5+.85x is less than or equal to 15
I have a more better explaination on paper.
ANSWER:5

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Given the rhombus ABCD, find the area. Show all the work
dimaraw [331]

Answer:

Step-by-step explanation:

Diagonal AC=6

Diagonal BD=8

area=(AC×BD)/2=(6×8)/2=24 sq. units

6 0
3 years ago
From a point at the waters edge, a bridge arch 53 m away has an angle of elevation of 30, and the top of a yacht mast 14 m away
Arte-miy333 [17]

Answer:

Yes

Step-by-step explanation:

We don't  even have to do a calculation.  

The arch of the bridge is at point C.

The top of the yacht's mast is at E.

Even though the yacht is much closer to the water's edge, its mast has a much smaller angle of elevation.

The yacht can easily pass under the bridge.

3 0
3 years ago
If each commercial lasts 1/2 minute how many commercials are there in a 4-hour program?
Ivenika [448]

Answer:

if each commercial lasts 1/2 minute how many commercials are there in a 4-hour program?

480

Step-by-step explanation:

4 hours= 240 mins

240/0.5=480

6 0
3 years ago
Which of the following statements best describes the graph of x + y =2?
andre [41]

Step-by-step explanation:

so x+y=2

a line would have to go 0,2 2,0

so the answer would be D

4 0
3 years ago
The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation
fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

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 question, we have that:

\mu = 197.5, \sigma = 8.3

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

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

Z = \frac{210.9 - 197.5}{8.3}

Z = 1.61

Z = 1.61 has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

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

By the Central Limit Theorem

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

Z = \frac{196 - 197.5}{2.14}

Z = -0.7

Z = -0.7 has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0
3 years ago
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