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2 years ago

15

I REALLY NEED YOUR HELP PLEASEEE ASAAAAPPPPP THE QUESTION GOES LIKE THIS, if bob has 1 apple and jimmy has 1 apple what will be

the total of all the apples? PLESE HELLPPPPP

2 answers:

Grace [21]2 years ago

8 0

Answer:

the apple will be 1

.............

Kazeer [188]2 years ago

3 0

Answer:

2

Step-by-step explanation:

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Tom rolls 2 fair dice and adds the results from each.

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Answer:

1/6

Step-by-step explanation:

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2 years ago

Help please quickly thank you so much

STALIN [3.7K]

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2 years ago

Read 2 more answers

The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated b

storchak [24]

Answer:

A task time of 177.125s qualify individuals for such training.

Step-by-step explanation:

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

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

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

A distribution that can be approximated by a normal distribution with a mean value of 145 sec and a standard deviation of 25 sec, so $\mu = 145, \sigma = 25$ .

The fastest 10% are to be given advanced training. What task times qualify individuals for such training?

This is the value of X when Z has a pvalue of 0.90.

Z has a pvalue of 0.90 when it is between 1.28 and 1.29. So we want to find X when $Z = 1.285$ .

So

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

$1.285 = \frac{X - 145}{25}$

$X - 145 = 32.125$

$X = 177.125$

A task time of 177.125s qualify individuals for such training.

7 0

3 years ago

If you answer this you will get 69 points but please answer asap

spin [16.1K]

Answer:

I think the answer for the first one is 3 I am not sure about the second one

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2 years ago

2.5% of a population are infected with a certain disease. There is a test for the disease, however the test is not completely ac

Aleks04 [339]

Answer:

$P(disease/positivetest) = 0.36116$

Step-by-step explanation:

This is a conditional probability exercise.

Let's name the events :

I : ''A person is infected''

NI : ''A person is not infected''

PT : ''The test is positive''

NT : ''The test is negative''

The conditional probability equation is :

Given two events A and B :

P(A/B) = P(A ∩ B) / P(B)

$P(B) >0$

P(A/B) is the probability of the event A given that the event B happened

P(A ∩ B) is the probability of the event (A ∩ B)

(A ∩ B) is the event where A and B happened at the same time

In the exercise :

$P(I)=0.025$

$P(NI)= 1-P(I)=1-0.025=0.975\\P(NI)=0.975$

$P(PT/I)=0.904\\P(PT/NI)=0.041$

We are looking for P(I/PT) :

P(I/PT)=P(I∩ PT)/ P(PT)

$P(PT/I)=0.904$

P(PT/I)=P(PT∩ I)/P(I)

0.904=P(PT∩ I)/0.025

P(PT∩ I)=0.904 x 0.025

P(PT∩ I) = 0.0226

P(PT/NI)=0.041

P(PT/NI)=P(PT∩ NI)/P(NI)

0.041=P(PT∩ NI)/0.975

P(PT∩ NI) = 0.041 x 0.975

P(PT∩ NI) = 0.039975

P(PT) = P(PT∩ I)+P(PT∩ NI)

P(PT)= 0.0226 + 0.039975

P(PT) = 0.062575

P(I/PT) = P(PT∩I)/P(PT)

$P(I/PT)=\frac{0.0226}{0.062575} \\P(I/PT)=0.36116$

8 0

2 years ago