4

To examine the potential effects on the transmission of dengue, Bian et al. (2010) infected mosquitoes from both the WB1 (labora

nekit [7.7K]

Answer:

reject H0

the viral titer is lower in WB1

Step-by-step explanation:

yeah

6 0

4 years ago

A Department of Transportation report about air travel found that nationwide, 76% of all flights are on time. Base the following

ser-zykov [4K]

Answer:

30 flights are expected to be late.

Step-by-step explanation:

Consider the provided information.

A Department of Transportation report about air travel found that nationwide, 76% of all flights are on time.

That means 100-76% = 24% of all flights are not on time.

125 randomly selected flights.

We need to find flights would you expect to be late.

Flight expect to be late E(x) = nq

Here n is 125 and the probability of late is 24 or q = 0.24

Thus substitute the respective values in the above formula.

Flight expect to be late E(x) = 125 × 0.24 = 30

Hence, the 30 flights are expected to be late.

5 0

3 years ago

A sphere has a radius 14 units. What is the surface area of the sphere

morpeh [17]

Answer:

2463.009

radius squared x 4 x pi is the formula

14 x 14 = 196 x 4 = 784 x pi = 2463.00864

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

The results of a national survey showed that on average, adults sleep 6.7 hours per night. Suppose that the standard deviation i

Roman55 [17]

Answer:

a) Atleast 75%

b) Atleast 88.9%

c) 95%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 6.7

Standard Deviation, σ = 1.6

Chebyshev's Theorem:

Atleast $1 - \dfrac{1}{k^2}$ percent of data lies within k standard deviation of mean.

Empirical Rule:

Almost all the data lies within 3 standard deviation for a normal data.
Around 68% of data lies within 1 standard deviation of mean.
Around 95% of data lies within 2 standard deviation of mean.
Around 99.7% of data lies within 3 standard deviations o mean.

a) minimum percentage of individuals who sleep between 3.5 and 9.9 hours

$3.5 = \mu - 2\sigma = 6.7 - 2(1.6)\\9.9 = \mu + 2\sigma = 6.7 + 2(1.6)$

We have to find the percent of data lying within 2 standard deviation of mean.

Putting k = 2

$1 - \dfrac{1}{(2)^2} = 0.75 = 75\%$

Atleast 75% of of individuals who sleep between 3.5 and 9.9 hours.

b) minimum percentage of individuals who sleep between 2.7 and 10.7 hours

$2.7 = \mu - 3\sigma = 6.7 - 3(1.6)\\10.7 = \mu + 3\sigma = 6.7 + 3(1.6)$

We have to find the percent of data lying within 3 standard deviation of mean.

Putting k = 3

$1 - \dfrac{1}{(3)^2} = 0.889 = 88.9\%$

Atleast 88.9% of of individuals who sleep between 2.7 and 10.7 hours.

c) percentage of individuals who sleep between 3.5 and 9.9 hours per day.

Thus, we have to find the percentage of data lying within 2 standard deviation.

By empirical rule, around 95% of data lies within 2 standard deviation.

Thus, 95% of of individuals who sleep between 3.5 and 9.9 hours.

This is higher than the percentage obtained from Chebyshev's rule.

4 0

3 years ago

Solve the equation 2/5x-1=9. Complete each step of the solution and check your solution

Fofino [41]

Answer:

x = 25

Step-by-step explanation:

2/5x - 1 = 9

Add 1 to both sides

2/5x = 10

Multiply by 5 on both sides

2x = 50

Divide by 2 on both sides

x = 25

Check the answer:

Plug in 25 for x

2/5(25) - 1 = 9

10 - 1 = 9

9 = 9

The answer is correct.

3 0

3 years ago