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LekaFEV [45]
3 years ago
9

HELP PLS ASAP PLSSSSSs

Mathematics
1 answer:
emmasim [6.3K]3 years ago
8 0

Answer:

6 companies.

Step-by-step explanation:

[When you read a stem and leaf plot, the lowest number in the stem category, like 3 for example, would be in the tens place. The numbers in the leaf category are the ones places.]

"At least" tells you that anything equal to or higher the number given is relevant to the question. If we count the amount of numbers that are 63 or over, we have:

63, 69, 70, 73, 76, 78.

These are 6 numbers.

Your answer would be <em>6</em> companies.

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A company makes rubber rafts. 12% of them develop cracks within the first month of operation. 27 new rafts are randomly sampled
rewona [7]

The probability that the number of tested rafts that develop cracks is no more than 3 is <u>.00006</u>.

The true proportion, p for the population is given to 0.12.

Thus, the mean, μ, for the sample = np = 27*0.12 = 3.24.

The sample size, n, given to us is 27.

Thus, the standard deviation, s, for the sample can be calculated using the formula, s = √{p(1 - p)}/n.

s = √{0.12(1 - 0.12)}/27 = √0.003911 = 0.0625389.

We are asked to calculate the probability that the number of tested rafts that develop cracks is no more than 3, that is, we need to calculate P(X ≤3).

P(X ≤ 3)

= P(Z ≤ {(3 - 3.24)/0.0625389) {Using the formula z = (x - μ)/s}

= P(Z ≤ -3.8376114706)

= .00006 {From table}.

Thus, the probability that the number of tested rafts that develop cracks is no more than 3 is <u>.00006</u>.

Learn more about sampling distributions at

brainly.com/question/15507495

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8 0
1 year ago
Help with questions 21-23 using pemdas
taurus [48]

Answer:

21. 123    22. 6.86 or 6/ 6/7  23. 90

Step-by-step explanation:

21. (6+4)² + (22+10÷2)

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22. (11+42-5) ÷ (11-4)

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Vesna [10]

You're looking for the largest number <em>x</em> such that

<em>x</em> ≡ 1 (mod 451)

<em>x</em> ≡ 4 (mod 328)

<em>x</em> ≡ 1 (mod 673)

Recall that

<em>x</em> ≡ <em>a</em> (mod <em>m</em>)

<em>x</em> ≡ <em>b</em> (mod <em>n</em>)

is solvable only when <em>a</em> ≡ <em>b</em> (mod gcd(<em>m</em>, <em>n</em>)). But this is not the case here; with <em>m</em> = 451 and <em>n</em> = 328, we have gcd(<em>m</em>, <em>n</em>) = 41, and clearly

1 ≡ 4 (mod 41)

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So there is no such number.

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Answer:

43,81

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