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3 years ago

6

the odds of rolling a certain number on a single die is 1 chance in 6 rolls in the course of a game someone rolled 3 fives accor

ding to the odds how many times must that person have thrown the die?

2 answers:

monitta3 years ago

5 0

Answer:

18

Step-by-step explanation:

3 = 1/6 × n

n = 18

sergeinik [125]3 years ago

3 0

108 times. That is the probability of that happening.

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3 years ago

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If x^2 - 16 > 0, which of the following must be true?

zimovet [89]

The left hand side of the equation is a difference of two squares and may be factored out as follows,
(x - 4)(x + 4) > 0
They may be individually taken as,
x - 4 > 0 ; x > 4
x + 4 > 0 ; x < -4
Thus, the answer to this item is letter A.

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Engineers must consider the diameters of heads when designing helmets. The company researchers have determined that the populati

Margaret [11]

Answer:

1. The minimum head breadth that will fit the clientele is of 4.41-in.

2. The maximum head breadth that will fit the clientele is of 7.19-in.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions 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.

Normally distributed with a mean of 5.8-in and a standard deviation of 0.8-in.

This means that $\mu = 5.8, \sigma = 0.8$

1. What is the minimum head breadth that will fit the clientele?

The 4.1st percentile, that is, X when Z has a pvalue of 0.041, so X when Z = -1.74.

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

$-1.74 = \frac{X - 5.8}{0.8}$

$X - 5.8 = -1.74*0.8$

$X = 4.41$

The minimum head breadth that will fit the clientele is of 4.41-in.

2. What is the maximum head breadth that will fit the clientele?

100 - 4.1 = 95.9th percentile, that is, X when Z has a pvalue of 0.959, so X when Z = 1.74.

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

$1.74 = \frac{X - 5.8}{0.8}$

$X - 5.8 = 1.74*0.8$

$X = 7.19$

The maximum head breadth that will fit the clientele is of 7.19-in.

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3 years ago

ms.vellejo has these cans of soup in her cabinet 1 can of tomato soup 4 cans of chicken soup 3 cans of cheese soup 2 cans of bee

Flauer [41]

Answer:

There is a 40% chance that she will choose a can of beef soup, a 20% chance that she will choose a can of chicken soup, and a combined 60% chance that she will choose a can of beef or chicken soup.

Step-by-step explanation:

Since Ms. Vellejo has 1 can of tomato soup, 4 cans of chicken soup, 3 cans of cheese soup and 2 cans of beef soup in her cabinet, and she will randomly choose one can of soup, to determine what is the probililty that she will choose a can of beef or chicken soup, the following calculation must be performed:

1 + 4 + 3 + 2 = 10

Beef soup = 2 cans = 2/10 = 0.2 = 20%

Chicken soup = 4 cans = 4/10 = 0.4 = 40%

20 + 40 = 60

Thus, there is a 40% chance that she will choose a can of beef soup, a 20% chance that she will choose a can of chicken soup, and a combined 60% chance that she will choose a can of beef or chicken soup.

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3 years ago