Helppppppppppppp0ppppppp

Each pitcher of power smoothie that Ginger makes jas 2 scoops of pineapple, 3 scoops of strawberries, 1 scoop of spinach, and 1

Virty [35]

Answer:

56 scoops

Step-by-step explanation:

.

8 0

3 years ago

Simplify the expression<br> –2 + 7(x + 8)

anastassius [24]

Answer:

54+7x

Step-by-step explanation:

-2+7(x+8)

-2+7x+56

54+7x

7 0

3 years ago

Read 2 more answers

lbvjy [14]

$\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&Sides&Area&Volume\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array} \\\\
-----------------------------\\\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
-------------------------------\\\\$

$\bf \cfrac{smaller}{larger}\qquad \cfrac{s}{s}=\cfrac{\sqrt{48\pi }}{\sqrt{75\pi }}\implies \cfrac{s}{s}=\cfrac{\sqrt{(2^2)^2\cdot 3}}{\sqrt{5^2\cdot 3}}\implies \cfrac{s}{s}=\cfrac{4\sqrt{3}}{5\sqrt{3}}
\\\\\\
\cfrac{s}{s}=\cfrac{4}{5}$

8 0

3 years ago

Read 2 more answers

Heights of men have a bell-shaped distribution, with a mean of 176 cm and a standard deviation of 7 cm. Using the Empirical Rule

Vaselesa [24]

Answer:

a) 68% of the men fall between 169 cm and 183 cm of height.

b) 95% of the men will fall between 162 cm and 190 cm.

c) It is unusual for a man to be more than 197 cm tall.

Step-by-step explanation:

The 68-95-99.5 empirical rule can be used to solve this problem.

This values correspond to the percentage of data that falls within in a band around the mean with two, four and six standard deviations of width.

<em>a) What is the approximate percentage of men between 169 and 183 cm? </em>

To calculate this in an empirical way, we compare the values of this interval with the mean and the standard deviation and can be seen that this interval is one-standard deviation around the mean:

$\mu-\sigma=176-7=169\\\mu+\sigma=176+7=183$

Empirically, for bell-shaped distributions and approximately normal, it can be said that 68% of the men fall between 169 cm and 183 cm of height.

<em>b) Between which 2 heights would 95% of men fall?</em>

This corresponds to ±2 standard deviations off the mean.

$\mu-2\sigma=176-2*7=162\\\\\mu+2\sigma=176+2*7=190$

95% of the men will fall between 162 cm and 190 cm.

<em>c) Is it unusual for a man to be more than 197 cm tall?</em>

The number of standard deviations of distance from the mean is

$n=(197-176)/7=3$

The percentage that lies outside 3 sigmas is 0.5%, so only 0.25% is expected to be 197 cm.

It can be said that is unusual for a man to be more than 197 cm tall.

3 0

3 years ago

On any day, the probability of rain is 0.3. The occurrence of rain on any day is independent of the occurrence of rain on any ot

poizon [28]

<h3>Answer: 0.47178 Step-by-step explanation: Find the probability for each p(X=x) up to 5 using the equation: (x-1)C(r-1)*p^r * q^x-r, where x is number of days, p = .3 (prob of rain). q=.7 (prob of not rain), and r=2 (second day of rain). also C means choose. So p(X=1) = 0 p(X=2) = 1C1 * .3^2 * .7^0 = .09 P(X=3) = 2C1 * .3^2 * .7^1 = .126 P(X=4) = 3C1 * .3^2 * .7^2 = .1323 P(X=5) = 4C1 * .3^2 * .7^3 = .12348 Then add all of them up 0+.09+.126+.1323+.12348 = .47178</h3>

3 0

3 years ago