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UNO [17]
3 years ago
6

Cesar, Carmen, and Dalila raised $95.34 for their tennis team. Carmen raised $12.12 less than Cesar, and Cesar raised $35 more t

han Dalila. If X = the amount raised by Carmen, choose the expressions that represent the amount each other player raised. Check All that Apply. Thank you in advance.
A.) x + 47.12 B.) x + 12.12 C.) 2x + 12.12 D.) x - 22.88
Mathematics
1 answer:
NeTakaya3 years ago
5 0

Answer:

Cesar's amount can be expressed by x + 12.12

Dalila's amount can be expressed as  x - 22.88

Step-by-step explanation:

If Carmen raised 12.12.less than Cesar, that means his amount is 12.12 more than hers so that is <u>x + 12.12</u> for <u>Cesar</u>.

If  Cesar raised 35 more than Dalila, her amount is Cesar's minus 35. Calculate: x+12.12 - 35. the difference is 22.88 less than x  so the amount for <u>Dalila is x- 22.88</u>

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Answer:

  • B. 5sqrt(2)

Step-by-step explanation:

  • x² = 4² + (√34)²
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8 0
3 years ago
Read 2 more answers
A: 28, 23,30, 25, 27
Serggg [28]

Answer:

8

Step-by-step explanation:

mean of A=28+23+30+25+27/5 = 26.6

average of B=22+19+15+17+20/5 = 18.6

now,

difference= 26.6-18.6

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6 0
3 years ago
An article written for a magazine claims that 78% of the magazine's subscribers report eating healthily the previous day. Suppos
Schach [20]

Answer:

89.44% probability that less than 80% of the sample would report eating healthily the previous day

Step-by-step explanation:

We use the binomial approximation to the normal 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)}.

In this problem, we have that:

p = 0.78, n = 675

So

\mu = E(X) = np = 675*0.78 = 526.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{675*0.78*0.22} = 10.76

What is the approximate probability that less than 80% of the sample would report eating healthily the previous day?

This is the pvalue of Z when X = 0.8*675 = 540. So

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

Z = \frac{540 - 526.5}{10.76}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

89.44% probability that less than 80% of the sample would report eating healthily the previous day

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4 years ago
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Check the attached file and let me know what you think.
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4 years ago
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