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valentinak56 [21]
3 years ago
13

Does anyone know the answer to this?

Mathematics
2 answers:
ivolga24 [154]3 years ago
5 0
It’s
Y=5/6x-8 and 5/6x+4
Rainbow [258]3 years ago
4 0

Answer:

the last one

Step-by-step explanation:

Two lines will be parallel when their slopes are equal, and two lines will be perpendicular when their slopes are negative reciprocals of each other. Our slopes for these two equations are the coefficient for the x value. Both slopes are equal so these lines are parallel.

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Can someone answer this its just boxplots :)))
Kobotan [32]

Answer:

B.

Step-by-step explanation:

The answer is B because the box and whisker plot has a line in the middle to show the mean or average. C may seem correct but it is not because the range is actually 100-900.

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Need help with this ASAP!!
postnew [5]

Answer:

64

Step-by-step explanation:

40x1.6

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A factory produces 1,250,000 toys each year. The number of toys is expected to increase by about 150% per year.
Marizza181 [45]

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The correct option is D. n= 1.25(2.5^t)

Step-by-step explanation:

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6 0
3 years ago
Destiny said that 0.6 divided by 2 equals 0.3 is she correct​
Reil [10]

Answer:

Yes

Step-by-step explanation:

8 0
3 years ago
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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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