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3 years ago

Hear is a poem

Mathematics

2 answers:

Alex73 [517]3 years ago

8 0

Answer: COOOOOOOOOOOOOL

Mumz [18]3 years ago

4 0

Answer:

wew so wha do you need help withhhhhhh

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A bowl contains 13 red and brown M&Ms. There are 3 more red M&Ms than brown. How many of each color in a bowl. I have to

maks197457 [2]

Answer:

So there are 8 red and 5 brown M&M

Step-by-step explanation:

Let X is red M&M

Let Y is brown M&M

We know that:

3 more red M&Ms than brown <=> X = Y +3
A bowl contains 13 red and brown M&Ms <=> X + Y = 13

substitute X into the second equation: (Y + 3) + Y = 13

<=> 2Y = 10

<=> Y =5

=> X = 5 + 3 = 8

So there are 8 red and 5 brown M&M

8 0

4 years ago

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(2^3)^-1(4m^3)^2<br><br>^ means exponent

Elden [556K]

Wouldnt -128m^6 be the answer if there is no value for m?

8 0

3 years ago

Look at the picture for the question please show work.

lukranit [14]

$A=\dfrac{1}{2}ah\\
a=25 \text{ mm}\\
h=14 \text{ mm}\\\\
A=\dfrac{1}{2}\cdot25\cdot14\\
A=175 \text{ mm}^2 \Rightarrow C$

6 0

4 years ago

The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to d

k0ka [10]

Part A:

From the central limit theorem, since the number of samples is large enough (up to 30), the mean of the the mean of the average number of moths in 30 traps is 0.6.

Part B:

The standard deviation is given by the population deviation divided by the square root of the sample size.

$standard \ deviation= \frac{\sigma}{\sqrt{n}} = \frac{0.4}{\sqrt{30}} = \frac{0.4}{5.477} =0.073$

Part C:

The probability that an approximately normally distributed data with a mean, μ, and the standard deviation, σ, with a sample size of n is greater than a number, x, given by

$P(X\ \textgreater \ x)=1-P(X\ \textless \ x) \\ \\ =1-P\left(z\ \textless \ \frac{x-\mu}{\sigma/\sqrt{n}} \right)$

Thus, given that the mean is 0.6 and the standard deviation is 0.4, the probability that <span>the average number of moths in 30 traps is greater than 0.7</span> given by:

$P(X\ \textgreater \ 0.7)=1-P(X\ \textless \ 0.7) \\ \\ =1-P\left(z\ \textless \ \frac{0.7-0.6}{0.073} \right)=1-P(z\ \textless \ 1.369) \\ \\ =1-0.91455=\bold{0.0855}$

7 0

4 years ago

Solve pls brainliest

Natalija [7]

Answer:

1.8333

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers