The best thing to do is to find 10%, and you can do this by dividing $270 by 10, and therefore, $27 is equal to 10%. To find 40%, you've got to multiply $27 by 4, and this gives you $108. Therefore, gabriel saved $108 last month :)
hope this helps u

7 0

2 years ago

Normal Distribution. Cherry trees in a certain orchard have heights that are normally distributed with mu = 112 inches and sigma

Lubov Fominskaja [6]

Answer:

The probability that a randomly chosen tree is greater than 140 inches is 0.0228.

Step-by-step explanation:

Given : Cherry trees in a certain orchard have heights that are normally distributed with $\mu = 112$ inches and $\sigma = 14$ inches.

To find : What is the probability that a randomly chosen tree is greater than 140 inches?

Solution :

Mean - $\mu = 112$ inches

Standard deviation - $\sigma = 14$ inches

The z-score formula is given by, $Z=\frac{x-\mu}{\sigma}$

Now,

$P(X>140)=P(\frac{x-\mu}{\sigma}>\frac{140-\mu}{\sigma})$

$P(X>140)=P(Z>\frac{140-112}{14})$

$P(X>140)=P(Z>\frac{28}{14})$

$P(X>140)=P(Z>2)$

$P(X>140)=1-P(Z$

The Z-score value we get is from the Z-table,

$P(X>140)=1-0.9772$

$P(X>140)=0.0228$

Therefore, the probability that a randomly chosen tree is greater than 140 inches is 0.0228.

5 0

2 years ago

Please help me match this

Setler [38]

4/5=0.8
0.35=7/20
3/8=0.375
0.775=31/40

You could have use a calculator for this but you’re welcome

8 0

2 years ago