Help this is algebra

If, for a sample in which the subjects are randomly chosen, the mean income is $45,000, the sample size is 1600, the standard de

jeka57 [31]

Just add 45'000, 4,000, 10,divided by 95% = your answer.

6 0

4 years ago

A product you carry has a shelf life of 90 days. You purchased it September 5. Is it still good December 31?

Vitek1552 [10]

Answer:

no, 90 days after september 5 is december 4

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

The mean incubation time of fertilized chicken eggs kept at 100.5 degree F in a still-air incubator is 21 days. Suppose that the

Nitella [24]

Given Information:

Mean incubation time = 21 days

Standard deviation of incubation time = 1 day

Required Information:

a) P(X < 20) = ?

b) P(X > 22) = ?

c) P(19 < X < 21) = ?

Answer:

a) P(X < 20) = 15.87

b) P(X > 22) = 15.87

c) P(19 < X < 21) = 47.23%

Explanation:

What is Normal Distribution?

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.

a) We want to find out the probability that a randomly selected fertilized chicken egg hatches in less than 20 days.

$P(X < 20) = P(Z < \frac{x - \mu}{\sigma} )\\\\P(X < 20) = P(Z < \frac{20 - 21}{1} )\\\\P(X < 20) = P(Z < \frac{-1}{1} )\\\\P(X < 20) = P(Z < -1)\\$

The z-score corresponding to -1 is 0.1587

$P(X < 20) = 0.1587\\\\P(X < 20) = 15.87 \%$

Therefore, the probability that a randomly selected fertilized chicken egg hatches in less than 20 days is 15.87%

b) We want to find out the probability that a randomly selected fertilized chicken egg takes over 22 days to hatch?

$P(X > 22) = 1 - P(X < 22)\\\\P(X > 22) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\P(X > 22) = 1 - P(Z < \frac{22 - 21}{1} )\\\\P(X > 22) = 1 - P(Z < \frac{1}{1} )\\\\P(X > 22) = 1 - P(Z < 1)\\$

The z-score corresponding to 1 is 0.8413

$P(X > 22) = 1 - 0.8413\\\\P(X > 22) = 0.1587\\\\P(X > 22) = 15.87 \%$

Therefore, the probability that a randomly selected fertilized chicken egg takes over 22 days to hatch is 15.87%

c) We want to find out the probability that a randomly selected fertilized chicken egg hatches between 19 and 21 days?

$P(19 < X < 21) = P( \frac{x - \mu}{\sigma} < Z < \frac{x - \mu}{\sigma} )\\\\P(19 < X < 21) = P( \frac{19-21}{1} < Z < \frac{21 - 21}{1} )\\\\P(19 < X < 21) = P( \frac{-2}{1} < Z < \frac{0}{1} )\\\\P(19 < X < 21) = P( -2 < Z < 0 )\\$

The z-score corresponding to -2 is 0.0227 and 0 is 0.50

$P(19 < X < 21) = P( Z < 0 ) - P( Z < -2 ) \\\\P(19 < X < 21) = 0.50 - 0.0227 \\\\P(19 < X < 21) = 0.4723\\\\P(19 < X < 21) = 47.23 \%$

Therefore, the probability that a child spends more than 4 hours and less than 8 hours per day unsupervised is 47.23%

How to use z-table?

Step 1:

In the z-table, find the two-digit number on the left side corresponding to your z-score. (e.g 1.0, 2.2, 0.5 etc.)

Step 2:

Then look up at the top of z-table to find the remaining decimal point in the range of 0.00 to 0.09. (e.g. if you are looking for 1.00 then go for 0.00 column)

Step 3:

Finally, find the corresponding probability from the z-table at the intersection of step 1 and step 2.

4 0

4 years ago

Andy is building a model of a square pyramid for a class project. The side length of the square base is 11 inches and the slant

Wewaii [24]

Answer:

The surface area of the model pyramid is $451\ in^{2}$