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olga55 [171]
3 years ago
6

David is preparing for a dance competition. He needs to practice for 14 hours in 5 days. Which of the following statements show

that David is practicing according to the required rate?
A.
55 hours in 10 days
B.
67 hours in 12 days
C.
28 hours in 10 days
D.
22 hours in 15 days
Mathematics
2 answers:
schepotkina [342]3 years ago
6 0

Answer:

it's answer is C

hope it helps you

Elena L [17]3 years ago
5 0

Answer:

B is your answer

Step-by-step explanation:

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Rosa conducted a survey of 25 students' favorite types of music. The results are shown in the table. Of the 600 students in the
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7 0
3 years ago
if the sum of two numbers is four and the sum of their squares minus three times their product is 76, find the numbers
Sauron [17]

Answer:

Numbers =(6,-2 )

Step-by step explanation:

Given

the sum of two numbers is four and the sum of their squares minus three times their product is 76,

Let the first term =X

Second term will be=4-x

So

X^2+(4-x)^2-3x(4-x)=76

x^2+16+x^2-8x-12x+3x^2=76

5x^2-20x+16=76

5x^2-20x-60=0

5(x^2-4x-12)=0

So x^2-4x-12=0

x^2-(6-2)x-12=0

x^2-6x+2x-12=0

X(x-6)+2(x-6)=0

(X-6)(X+2)=0

X=6,-2

Let X=6 and -2 then numbers will be 6 and (4-6)=-2

6 0
3 years ago
Consider the probability that greater than 96 out of 153 DVDs will work correctly. Assume the probability that a given DVD will
ICE Princess25 [194]

Answer:

0.3594 = 35.94% probability that greater than 96 out of 153 DVDs will work correctly.

Step-by-step explanation:

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:

n = 153, p = 0.62

So

\mu = E(X) = np = 153*0.62 = 94.86

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{153*0.62*0.38} = 6

Probability that greater than 96 out of 153 DVDs will work correctly.

97 or more DVDs, which is 1 subtracted by the pvalue of Z when X = 97. So

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

Z = \frac{97 - 94.86}{6}

Z = 0.36

Z = 0.36 has a pvalue of 0.6406

1 - 0.6406 = 0.3594

0.3594 = 35.94% probability that greater than 96 out of 153 DVDs will work correctly.

3 0
3 years ago
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