What’s the answer for this

A house hunter on Long Island estimates that 20% of the available houses in her price range are in acceptable condition. Further

Otrada [13]

Answer:

36% probability that she will find an acceptable house in the first two weeks that she looks.

Step-by-step explanation:

For each house, there is a 20% probability of it being acceptable. And a 100-20 = 80% of not being acceptable

What is the probability that she will find an acceptable house in the first two weeks that she looks (round off to second decimal place)?

She only looks one house a week.

So this is the same as the probability of taking two or less weeks to find a house.

The are two outcomes

Finding an acceptable house in the first week, with 20% probability

Not finding an acceptable house in the first week, with 80% probability, and then finding an acceptable house in the second week, with 20% probability.

Probability:

$P = 0.2 + 0.8*0.2 = 0.36$

36% probability that she will find an acceptable house in the first two weeks that she looks.

4 0

3 years ago

How can you check your answer?

mars1129 [50]

Answer:

Option D is correct

Step-by-step explanation:

Multiply 7 by 3. then add 3 and make sure the answer is 24.

7 x 3 + 3 = 21 + 3 = 24

Hope this helps!

:)

3 0

4 years ago

Please telll me ASAP

erica [24]

the answer is y=-2x+5

4 0

3 years ago

Students in a representative sample of 69 second-year students selected from a large university in England participated in a stu

Serhud [2]

Answer:

95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].

Step-by-step explanation:

We are given that for the 69 second-year students in the study at the university, the sample mean procrastination score was 41.00 and the sample standard deviation was 6.89.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

P.Q. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean procrastination score = 41

s = sample standard deviation = 6.89

n = sample of students = 69

$\mu$ = population mean estimate

Here for constructing 95% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the true mean, $\mu$ is ;

P(-1.9973 < $t_6_8$ < 1.9973) = 0.95 {As the critical value of t at 68 degree

of freedom are -1.9973 & 1.9973 with P = 2.5%}

P(-1.9973 < $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ < 1.9973) = 0.95

P( $-1.9973 \times{\frac{s}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $1.9973 \times{\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.9973 \times{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.9973 \times{\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ =[ $\bar X-1.9973 \times{\frac{s}{\sqrt{n} } }$ , $\bar X+1.9973 \times{\frac{s}{\sqrt{n} } }$ ]

= [ $41-1.9973 \times{\frac{6.89}{\sqrt{69} } }$ , $41+1.9973 \times{\frac{6.89}{\sqrt{69} } }$ ]

= [39.34 , 42.66]

Therefore, 95% confidence interval estimate of μ, the mean procrastination scale for second-year students at this terval college is [39.34 , 42.66].

5 0

3 years ago

Which point below is not on the graph A.)(-13, 7) B.)(-35, 1) C.)(11, 5) D.)(27, 3)

denis-greek [22]

Answer:

It is B

Step-by-step explanation:

Because their isnt a square root fpr 51

4 0

3 years ago