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ira [324]
3 years ago
15

-1.2 = 1.8x what does x equal?

Mathematics
2 answers:
Eddi Din [679]3 years ago
6 0

Answer:

Step-by-step explanation:

. 6

Oksanka [162]3 years ago
4 0

Answer:

2/3 or .6666666666......

Step-by-step explanation:

isolate x  by dividing -1.2 by 1.8

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Giving brainliest *easy*!!
Norma-Jean [14]

Answer:

452.16 square yards

Step-by-step explanation:

The equation for the SA of a sphere is A=4πr2

Substituting r = 6 yards : A= 4 * 3.14 * 6^2 = 452.16 square yards

6 0
3 years ago
Read 2 more answers
Greatest common factor for 17 and 42
Musya8 [376]
Since 17 isnt a factor of 42 the only factor is 1
5 0
3 years ago
The chance of you having the same DNA as another person (other than an identical twin) is approximately 1 in
Sphinxa [80]

Answer:

10×10^-13

Hope this helps!

8 0
3 years ago
Please help me I’m super confused on this problem!
Lorico [155]

Answer:

m < amc = 54°

Step-by-step explanation:

< amb and < bmc are complementary angles whose sum equals 90°.

Therefore, to find the value of 2x°, we must first solve for x.

We can establish the following equality statement:

< amb + < bmc = < amc

< 2x° + (x + 9)° = 90°

Combine like terms:

2x° + x° + 9° = 90°

3x° + 9° = 90°

Subtract 9 from both sides:

3x° + 9° - 9° = 90° - 9°

3x = 81°

Divide both sides by 3 to solve for x:

3x/3 = 81°/3

x = 27°.

Since x = 27°, substitute its value into 2x° to find m < amc:

2x° = 2(27°) = 54°

Therefore, m < amc = 54°

Please mark my answers as the Brainliest, if you find this helpful :)

5 0
2 years ago
The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra
Snezhnost [94]
Let X denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by X_1,\ldots,X_{36}, each independently and identically distributed with distribution X_i\sim\mathcal N(26,7.2).

You want to find

\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)

Recall that if X\sim\mathcal N(\mu,\sigma), then the sampling distribution \overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right) with n being the size of the sample.

Transforming to the standard normal distribution, you have

Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}

so that in this case,

Z=6\dfrac{\overline X-26}{7.2}

and the probability is equivalent to

\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)
=\mathbb P(Z>1.481)\approx0.0693
5 0
3 years ago
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