All categories

3 years ago

Plsss help me with this NO LINKS PLSSS OR THEY WILL BE REPORTED and plss don't. do it just for the points

Mathematics

2 answers:

goldfiish [28.3K]3 years ago

6 0

I believe it would be 15 packages per week!!!

Sholpan [36]3 years ago

5 0

He would need to sell rougly 15 packages per week.

You might be interested in

Compare the two functions in the table. Will the value of function y = 2x eventually exceed the value of function y = x10?

Andrew [12]

First we rewrite the functions:
y = 2x
y = x ^ 10
We note that the second function always has values of y greater than the first function. However, there is a value of x for which the first function is greater.
For x = 1 we have:
y = 2 (1) = 2
y = (1) ^ 10 = 1
We note that:
2> 1
Answer:
Yes, the value of function y = 2x eventually exceed the value of function y = x ^ 10.

5 0

3 years ago

Read 2 more answers

If the class creates a histogram of the data in the table, how many students are in the range 12 cm to 13.9 cm?

muminat

Answer:

4 i think

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The chart below shows conversion between kilometers and miles.

Feliz [49]

Answer:

Step-by-step explanation:

1.2 divided by 2 is 0.6 and you want to know how much twenty is bc thats the whole point of the question lol

but anyway multiply 0.6 and 20 to get 12

6 0

3 years ago

Read 2 more answers

One measure of an athlete’s ability is the height of his or her vertical leap. Many professional basketball players are known fo

almond37 [142]

Answer:

(1) P( $\bar X$ < 26 inches) = 0.0436

(2) P(27.5 inches < $\bar X$ < 28.5 inches) = 0.2812

Step-by-step explanation:

We are given that the mean vertical leap of all NBA players is 28 inches. Suppose the standard deviation is 7 inches and 36 NBA players are selected at random.

Firstly, Let $\bar X$ = mean vertical leap for the 36 players

Assuming the data follows normal distribution; so the z score probability distribution for sample mean is given by;

Z = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean vertical leap = 28 inches

$\sigma$ = standard deviation = 7 inches

n = sample of NBA player = 36

(1) Probability that the mean vertical leap for the 36 players will be less than 26 inches is given by = P( $\bar X$ < 26 inches)

P( $\bar X$ < 26) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{26-28}{\frac{7}{\sqrt{36} } }$ ) = P(Z < -1.71) = 1 - P(Z $\leq$ 1.71)

= 1 - 0.95637 = 0.0436

(2) <em>Now, here sample of NBA players is 26 so n = 26.</em>

Probability that the mean vertical leap for the 26 players will be between 27.5 and 28.5 inches is given by = P(27.5 inches < $\bar X$ < 28.5 inches) = P( $\bar X$ < 28.5 inches) - P( $\bar X$ $\leq$ 27.5 inches)

P( $\bar X$ < 28.5) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{28.5-28}{\frac{7}{\sqrt{26} } }$ ) = P(Z < 0.36) = 0.64058 {using z table}

P( $\bar X$ $\leq$ 27.5) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{27.5-28}{\frac{7}{\sqrt{26} } }$ ) = P(Z $\leq$ -0.36) = 1 - P(Z < 0.36)