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3 years ago

13

A surgical technique is performed on eleven patients. You are told there is 90% chance of success. Find the probability that the

surgery is successful for at least 9 patients

1 answer:

Alexandra [31]3 years ago

3 0

Answer:

Your answer is : 0.7748

Step-by-step explanation:

Use bernoulli probability

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Elina [12.6K]

They have the same volume, so it's not a or d, but the surface area of a is 104, and the surface area of b is 94, so the answer is B.
Hope i helped! i also have this question in k12.<span />

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3 years ago

Perform the indicated row operation:

alexandr1967 [171]

Answer:

answers down below

Step-by-step explanation:

lol i hope it's this one

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2 years ago

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The Precision Scientific Instrument Company manufactures thermometers that are supposed to give readings of 0degrees°C at the fr

Sergio039 [100]

Answer:

Step-by-step explanation:

Hello!

The variable of study is X: Temperature measured by a thermometer (ºC)

This variable has a distribution approximately normal with mean μ= 0ºC and standard deviation σ= 1.00ºC

To determine the value of X that separates the bottom 4% of the distribution from the top 96% you have to work using the standard normal distribution:

P(X≤x)= 0.04 ⇒ P(Z≤z)=0.04

First you have to use the Z tables to determine the value of Z that accumulates 0.04 of probability. It is the "bottom" 0.04, this means that the value will be in the left tail of the distribution and will be a negative value.

z= -1.75

Now using the formula of the distribution and the parameters of X you have to transform the Z-value into a value of X

z= (X-μ)/σ

z*σ = X-μ

(z*σ)+μ = X

X= (-1.75-0)/1= -1.75ºC

The value that separates the bottom 4% is -1.75ºC

I hope this helps!

5 0

3 years ago

Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we

Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N( $\mu = 7 lbs , \sigma^{2} = 0.875^{2} lbs$ )

The standard normal z distribution is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, X bar = sample mean weight

n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{7.5-7}{\frac{0.875}{\sqrt{4} } }$ ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

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3 years ago

5. $120 for 15 books A. $8−−−−1 book B. 1 book−−−−$8 C. $15−−−−1 book D. 1 book−−−−$15

ddd [48]

Answer:

it should be A and if it is not a then b but I think A

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3 years ago

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