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3 years ago

PLEASE HELP !! WILL MARK BRAINLIEST TO WHOEVER GETS IT RIGHT !!

Mathematics

1 answer:

Oliga [24]3 years ago

8 0

Answer: 2

Step-by-step explanation:

Sub in 2 to both equations

$2^2-1=-2+5\\4-1=3\\3=3$

As you see we see that both sides result into 3 (which would be the y) meaning they both intersect when x=2 and y=3

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Josh earned $80 to use for back-to-school shopping. So far he has spent $24 on a backpack and $32 on new shoes. Think abt the yo

Lady_Fox [76]

Answer:

24$

Step-by-step explanation:

$80 - $32 - $24 = $24

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3 years ago

Help please!! I'm having trouble with this question

Fiesta28 [93]

Answer:

the least value would be 85

Step-by-step explanation:

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3 years ago

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What is my number my number is a perfect square the only prime number in it prime factorization is 2 my number is a factor of 32

WITCHER [35]

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4 years ago

A where and a cylinder have the same radius and height . The volume of the cylinder is 48 cm 3. What is the volume of the sphere

SOVA2 [1]

Given:

Sphere and cylinder have same radius and height.

Volume of the cylinder = 48 cm³

To find:

The volume of the sphere.

Solution:

Radius and height of cylinder are equal.

⇒ r = h

Volume of cylinder:

$V=\pi r^2h$

Substitute the given values.

$48=\pi r^2r$ (since r = h)

$48=\pi r^3$

$48=3.14 \times r^3$

Divide by 3.14 on both sides.

$$\frac{48}{3.14} =\frac{3.14\times r^3}{3.14}$

$$15.28=r^3$

Taking cube root on both sides, we get

2.48 = r

The radius of the cylinder is 2.48 cm.

Sphere and cylinder have same radius and height.

Volume of sphere:

$$V=\frac{4}{3} \pi r^3$

$$V=\frac{4}{3} \times 3.14 \times (2.48)^3$

$V=63.85$

The volume of the sphere is 63.85 cm³.

6 0

4 years ago

A truck loaded with 8000 electronic circuit boards has just pulled into a firm’s receiving dock. The supplier claims that no mor

Juliette [100K]

Answer:

The 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

Step-by-step explanation:

Let X = number of boards that fall outside the most rigid level of industry performance specifications.

In a random sample of 300 boards the number of defective boards was 12.

Compute the sample proportion of defective boards as follows:

$\hat p =\frac{12}{300}=0.04$

The (1 - α)% confidence interval for population proportion p is:

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

The critical value of z for 95% confidence level is,

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use a z-table.

Compute the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification as follows:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.04\pm1.96\sqrt{\frac{0.04(1-0.04)}{300}}\\=0.04\pm0.022\\=(0.018, 0.062)\\\approx(1.8\%, 6.2\%)$

Thus, the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

3 0

3 years ago