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2 years ago

6x – 3y = 9 Solve for y

Mathematics

1 answer:

viktelen [127]2 years ago

7 0

Answer:

y=-3+2x

Step-by-step explanation:

Overall, move 6x to the right and then solve from there :).

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Point J is on line segment IK. Given JK = 8 and IJ = 7, determine the length IK.

masha68 [24]

Answer:

Step-by-step explanation:

If the line segment is IK, that means IJ+JK=IK. So, 7+8=15

3 0

2 years ago

(a) Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the prop

zalisa [80]

Answer:

(a) The sample sizes are 6787.

(b) The sample sizes are 6666.

Step-by-step explanation:

(a)

The information provided is:

Confidence level = 98%

MOE = 0.02

n₁ = n₂ = n

$\hat p_{1} = \hat p_{2} = \hat p = 0.50\ (\text{Assume})$

Compute the sample sizes as follows:

$MOE=z_{\alpha/2}\times\sqrt{\frac{2\times\hat p(1-\hat p)}{n}$

$n=\frac{2\times\hat p(1-\hat p)\times (z_{\alpha/2})^{2}}{MOE^{2}}$

$=\frac{2\times0.50(1-0.50)\times (2.33)^{2}}{0.02^{2}}\\\\=6786.125\\\\\approx 6787$

Thus, the sample sizes are 6787.

(b)

Now it is provided that:

$\hat p_{1}=0.45\\\hat p_{2}=0.58$

Compute the sample size as follows:

$MOE=z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})}{n}$

$n=\frac{(z_{\alpha/2})^{2}\times [\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})]}{MOE^{2}}$

$=\frac{2.33^{2}\times [0.45(1-0.45)+0.58(1-0.58)]}{0.02^{2}}\\\\=6665.331975\\\\\approx 6666$

Thus, the sample sizes are 6666.

7 0

2 years ago

Pls help me calculate the total area of compound shapes?!?!?

Elza [17]

Answer:

Left: 42ft^2

Right: 21ft^2

Step-by-step explanation:

Its the same question as last time.

Left: (3*10) + (2*6) = 42

Right: (3*5) + (2*3) = 21

4 0

1 year ago

In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. Furthermore, there

Vedmedyk [2.9K]

Answer:

a) 151lb.

b) 6.25 lb

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .