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kramer
3 years ago
5

Look at the graph shown below:

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
3 0

Answer:

A

Step-by-step explanation:

y=3/4x+4 best represents the line shown below. Because the slope of this graph is 3/4, and the y-intercept is positive 4. Remember that slope=rise/run. You can find the slope very easily by counting, 3 upwards and 4 rightwards.

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N² + 14 = 12n<br>how do I solve by quadratic formula ​
inysia [295]

Answer:

n = 6 + \sqrt{22}  or  n = 6 - \sqrt{22}

Step-by-step explanation:

We can solve this equation using the quadratic formula OR Completing the Square method.

n² + 14 = 12n

rearrange :  n² - 12n + 14  = 0  

here  a= 1 , b = -12,  c = 14

the quadratic formula says:   x =  - b/ (2a)  +  root(b^2 - 4ac) / (2a)

or  x =  - b/ (2a)  -  root(b^2 - 4ac) / (2a)

x =  - (-12)/ (2)  +  root((-12)^2 - 4*14) / (2)

x = 6  +  root (144 - 56) / 2

x = 6 + root(88)/2

x = 6 + root(4*22) / 2

x = 6 + 2*root(22)/2

x = 6 + root(22)  = 6 + \sqrt{22}

so   x =6 + \sqrt{22}   or  x = 6 - \sqrt{22}

In this case  x = n

n = 6 + \sqrt{22}  or  n = 6 - \sqrt{22}

3 0
3 years ago
Read 2 more answers
(2,2) (2,-4) (-5,-4)(-5,2)
grin007 [14]

Answer:

what's the question?

Step-by-step explanation:

5 0
2 years ago
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6. A sign in a bakery gives these options:
grigory [225]

Answer:

1. $0.48

2. $0.52

3. $0.53

Step-by-step explanation:

12/ $25

24/$46

50/$94

hope this helps

7 0
1 year ago
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Slove 4x - c = k for x
san4es73 [151]

Answer:

x=\frac{k+c}{4}

Step-by-step explanation:

5 0
2 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
2 years ago
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