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3 years ago

What is the best jumpshot in 2k20?

1 answer:

3 0

I'd say Isaiah Canaan or Kevin Love, Isaiah has quick reflexes and can keep up with the pace of the game, Kevin is basically assurance for anyone that they can land in the greenzone

You might be interested in

Alejandro needs to buy bird seed from the local pet store for his parrots. A

Gnesinka [82]

First take note he only has $20 and the bag he must buy is $12.55. Also the loose seeds are $1.49 per pound.

To solve subtract 12.55 from 20

20-12.55=7.45

So after buying the 10 lb bag he has $7.45 left now divide that by 1.49

7.45/1.49=5

So after buying the 10 lb bag he can also buy 5 lb of the loose seeds!

7 0

3 years ago

What is the slip of the line shown in the graph A) -2 B) -1 C) 1/-2 or D) 1/2

Serjik [45]

Answer:

m = -1/2

Step-by-step explanation:

$m=\frac{rise}{run}=\frac{1-2}{0+2}=\frac{-1}{2}=\boxed{-\frac{1}{2}}$

Hope this helps.

6 0

3 years ago

Write an equation for a horizontal line that passes that (4, -2)

NeTakaya

check the picture below.

8 0

3 years ago

A sandwich shop is ordering apples and grapes to make chicken salad.Apples cost $2.19 per pound and grapes cost $2.60 per pound.

andrew11 [14]

Answer:

Grapes=19.51

Step-by-step explanation:

Apples=$2.19 per pound

Grapes=$2.60 per pound

Apples+Grapes=20 pounds

A+G=20

A=20-G

Equation is

PaA+PgG=35.80

2.19A+2.60G=35.80

2.19(20-G)+2.60G=35.80

43.8-2.19G+2.60G=35.80

43.8-0.41G=35.80

-0.41G=35.80-43.8

-0.41G=-8

Divide both sides by 0.41

G=19.51 pounds

Recall

A=20-G

A=20-19.51

A=0.49 pounds

8 0

3 years ago

A small business owner estimates his mean daily profit as $970 with a standard deviation of $129. His shop is open 102 days a ye

Katena32 [7]

Answer:

The probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we select appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.

Then, the mean of the distribution of the sum of values of X is given by,

$\mu_{x}=n\mu$

And the standard deviation of the distribution of the sum of values of X is given by,

$\sigma_{x}=\sqrt{n}\sigma$

The information provided is:

μ = $970

σ = $129

n = 102

Since the sample size is quite large, i.e. n = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.

Then,

$\sum X\sim N(\mu_{x}=98940,\ \sigma_{x}=1302.84)$

Compute the probability that the shopkeeper's annual profit will not exceed $100,000 as follows:

$P (\sum X \leq 100,000) =P(\frac{\sum X-\mu_{x}}{\sigma_{x}}$

$=P(Z$

*Use a z-table for the probability.

Thus, the probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.

6 0

2 years ago