What’s the nth term of 7,14,23,34,47,62,79

In the last election, 72% of the voters

mezya [45]

Answer:

Option A) 19 to 25

Step-by-step explanation:

In the last election, 72% of the voters were in favor of a new outlet mall and

28% were against it.

Vanessa creates random samples of the voters using the numbers 1 to 25.

So, There are 25 numbers to represent all the voters

The range of number that could be used to represent the voters who were against the new outlet mall = 28% of 25 numbers = 0.28 * 25 = 7

So, 7 numbers only could be used to represent the voters who were against the new outlet mall.

Check the options to find which range has only 7 numbers.

A) 19 to 25 ⇒ 7 numbers

B) 7 to 18 ⇒ 12 numbers

C) 15 to 25 ⇒ 11 numbers

D) 1 to 18 ⇒ 18 numbers

The answer is option A) 19 to 25

6 0

3 years ago

A band has 36 members. They are arranged into 6 equal rows. How many band members are in each row? Can the same 36 members be pl

Galina-37 [17]

Answer:

6 people would be in each row // It isn't possible for there to be 7 equal rows

Step-by-step explanation:

36 members/6 rows= 6 people per row // 36 members/7 rows= 5.14..... people per row (just doesn't work!)

6 0

3 years ago

Find the length of the missing side when x=3. Round the answer to the nearest tenth.

Juliette [100K]

Answer/Step-by-step explanation:

The missing length can be found by applying pythagorean theorem. Thus:

Missing length = √((8x)² - (2x)²)

Missing length = √(64x² - 4x²)

✔️Missing length = √(60x²) = 2x√15

Plug in the value of x which is 3

Missing length = 2*3√15

✔️Length = 6√5

5 0

3 years ago

Simplify the expression: 4(2x-7) * 6x-7 08x-7 0 2x-28 8x-28

oksian1 [2.3K]

Answer:

Answer is (d)

Step-by-step explanation:

4(2x - 7)

8x - 28

3 0

3 years ago

Read 2 more answers

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard dev

Nesterboy [21]

Answer:

There is a 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

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:

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard deviation of 2.1-cm. This means that $\mu = 201.9, \sigma = 2.1$ .

For shipment, 9 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

By the Central Limit Theorem, since we are using the mean of the sample, we have to use the standard deviation of the sample in the Z formula. That is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{2.1}{\sqrt{9}} = 0.7$

This probability is 1 subtracted by the pvalue of Z when $X = 204.1$ .

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

$Z = \frac{204.1 - 201.9}{0.7}$

$Z = 3.14$

$Z = 3.14$ has a pvalue of 0.9992. This means that there is a 1-0.9992 = 0.0008 = 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

3 0

3 years ago