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Afina-wow [57]
2 years ago
14

Last week you worked 7 hours more than you normally do. Your typical week totals 36 hours.What was the total number of hours you

worked last week?
Mathematics
2 answers:
DENIUS [597]2 years ago
5 0
If your typical week totals to 36 hours, and it says you worked 7 hours more than you usually would, then that means you would just add 36 and 7, and you get your answer of 43 total hours.
 Hope this helped!
sergejj [24]2 years ago
4 0
The answer is 43 because 36 + 7 = 43
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Find the quotient of <br><br> -85^2-27n+76 / -17n-19
bagirrra123 [75]
<u><em>Answer:</em></u>
5n-4

<u><em>Explanation:</em></u>
<u>The given expression is:</u>
\frac{-85n^2-27n+76}{-17n-19}

<u>1- Take the negative as a common factor from both the numerator and the denominator. This will give us:</u>
\frac{-(85n^2+27n-76)}{-(17n+19)}

<u>2- Cancel out the negative sign (common factor) from the numerator and denominator. This will give us:</u>
\frac{(85n^2+27n-76)}{(17n+19)}

<u>3- Factor the numerator. This wil give:</u>
\frac{(5n-4)(17n+19)}{(17n+19)}

<u>4- Finally, cancel out the (17n+19) which common in both numerator and denominator. This will give us the final expression:</u>
5n-4

Hope this helps :)
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A certain medical test is known to detect 73% of the people who are afflicted with the disease Y. If 10 people with the disease
Setler [38]

The probability of an event is the measurement of the chance of that event's occurrence. The probabilities of considered events are:

  • P(At least 8 have the disease) ≈ 0.4378
  • P(At most 4 have the disease)  ≈ 0.0342
<h3 /><h3>How to find that a given condition can be modeled by binomial distribution?</h3>

Binomial distributions consist of n independent Bernoulli trials. Bernoulli trials are those trials that end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining to a binomial distribution with parameters n and p, then it is written as

X = B(n,p)

The probability that out of n trials, there'd be x successes is given by

P(X=x)  = ^nC_xp^x(1-p)^{n-x}

Since 10 people can be either diseased or not and they be so independent of each other (assuming them to be selected randomly) , thus, we can take them being diseased or not as outputs of 10 independent Bernoulli trials.

Let we say

Success= Probability of a diseased person tagged as diseased by the clinic

Failure = Probability of a diseased person tagged as not diseased by the clinic.

Then,

P(Success) = p = 72% = 0.72 (of a single person)

P(Failure) = q = 1-p = 0.28

Let X be the number of people diagnosed diseased by the clinic out of 10 diseased people. Then we have: X ≈ B(n+10,P=0.73)

Calculating the needed probabilities, we get:

a) P(At leased 8 have disease) = P(X≥8) =P(X=8) + P(X=9) + P(X=10)

P(X≥8) = ^{10}C_8(0.73)^8(0.28)^2+^{10}C_9(0.73)^9(0.27)^1+^{10}C_{10}(0.73)^{10}(0.27)^0

P(X≥8) ≈ 0.2548 + 0.1456 + 0.0374 ≈ 0.4378

b)  P(At most 4 have the disease) = P(X≤4) = P(X=0) + P(X=1)+P(X=2)+P(X=3)+P(X=4)

P (X ≤ 4) =

^{10}C_0(0.73)^0(0.27)^{10}+^{10}{C_1(0.73)^1(0.27)^9+^{10}{C_2(0.73)^2(0.27)^8+^{10}C_3(0.73)&^3(0.27)^7 \\

+^{10}C_4(0.73)^4(0.27)^6

P (X ≤ 4) = 0.000003 + 0.000076+0.00088+0.00604+0.02719

P (X ≤ 4) =  0.0342

Thus,

The probabilities of considered events are:

  • P(At leased 8 have disease) = 0.4378 approx
  • P(At most 4 have the disease)  = 0.0342 approx

Learn more about binomial distribution here:

brainly.com/question/13609688

#SPJ1

4 0
2 years ago
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