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Charra [1.4K]
2 years ago
7

What is the product of 2 numbers of 1 2 3 4 5 6 10 20 30​

Mathematics
1 answer:
Zanzabum2 years ago
5 0

Answer:

ekena la pernina de la abobo Kot Sa Rita kolos talas 8

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Solve the following probability problem. Give the answer in a
Lesechka [4]

Step-by-step explanation:

The probability of not selecting green or yellow marble is equal to the probability of selecting red and blue marble

n(S) =17

n(A)=5 (red)

n(B)=6 (blue)

answer is 5+6/17

=11/17

8 0
3 years ago
how to solve this prroblem by marianne buys 16 bags of potting soil that come in 5÷8_pound bags .how many pounds of potting soil
barxatty [35]
So, she buys 16 bags, each of them weights\frac{5}{8} pounds.

to find how many pounds of soil she bought, we have to multiply:

16*\frac{5}{8}=//16 is equal to 8*2, so we can simpify:
2*\frac{5}{1}

which is 2*5=10

so in the end she bought 10 pounds of soil!

3 0
3 years ago
In its Fuel Economy Guide for 2016 model vehicles, the Environmental Protection Agency gives data on 1170 vehicles. There are a
Nady [450]

Answer:

a. 19.68 miles per gallon.

b. 26.32 miles per gallon.

Step-by-step explanation:

Mean gas mileage (μ) = 23.0 mpg

Standard deviation (σ) = 4.9 mpg

In a normal distribution, for any length X, the z-score is determined by the following expression:

 z=\frac{X-\mu}{\sigma}

In a normal distribution, the 25th percentile  (first quartile) of a normal distribution has a corresponding z-score of z = -0.677 and the 75th percentile  has a corresponding z-score of z = 0.677

a. The first quartile of the distribution of gas mileage is

-0.677=\frac{X_{25}-23}{4.9}\\X_{25}=19.68\ mpg

19.68 miles per gallon.

b. The third quartile of the distribution of gas mileage is

0.677=\frac{X_{25}-23}{4.9}\\X_{25}=26.32\ mpg

26.32 miles per gallon.

5 0
3 years ago
Someone help me please
Bond [772]

Answer:

4

Step-by-step explanation:

Plug in the standard deviation formula. This is new to me but it took some research.. Hope it helps

6 0
3 years ago
PLEASE HELP!!!
Novosadov [1.4K]
\bf \qquad \qquad \textit{sum of an infinite geometric serie}
\\\\
S_n=\sum\limits_{i=1}^{\infty}\ a_1\cdot r^{i-1}\implies S=\cfrac{a_1}{1-r}\quad 
\begin{cases}
a_1=\textit{first term's value}\\
r=\stackrel{0\ \textless \ |r|\ \textless \ 1}{\textit{common ratio}}
\end{cases}

\bf \sum\limits_{n=1}^{\infty}~3\left(\frac{1}{4}  \right)^{n-1}~~
\begin{cases}
a_1=3\\
r=\frac{1}{4}
\end{cases}\implies S=\cfrac{3}{1-\frac{1}{4}}\implies S=\cfrac{\quad 3\quad }{\frac{3}{4}}
\\\\\\
S=\cfrac{\underline{3}}{1}\cdot \cfrac{4}{\underline{3}}\implies S=4
7 0
3 years ago
Read 2 more answers
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