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3 years ago

What’s the length of the hypotenuse in inches?

Mathematics

2 answers:

natta225 [31]3 years ago

5 0

Answer:

The hypotenuse is 25 inches.

Step-by-step explanation:

This is a right triangle, so the Pythagorean Theorem applies.

I will assume you know about the Pythagorean Theorem, therefore I will not define it.

Here, I am solving for the hypotenuse.

$a^2+b^2=c^2$

$24^2+7^2=c^2$

$576+49=c^2$

$625 = c^2$

$25=c$

ch4aika [34]3 years ago

4 0

Answer:

Step-by-step explanation:

To find the hypotenuse, you need to use a^2+b^2=c^2

Plugging the numbers in, the equation is

24^2+7^2=c^2

From here, you can just solve the equation, simplifying it equals

576+49=c^2

Add the two values

625=c^2

from here, you would square both sides to get c standing alone. This results in the equation finishing with the answer:

25=c

or flipping it around to make sence

c=25

Hope this helps! Have a good rest of your day!!!

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36^x-2=6 solve for x

Lina20 [59]

Answer:

x = $\frac{in (8)}{in (36)}$ if you consider doing decimal form x = 0.58027921. . .

Step-by-step explanation:

Take logarithm of both sides of the equation to remove the variable from the exponent.

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3 years ago

find the volume of a solid that is generated by rotating around the indicated axis the plane region bounded by the giveb curves:

Fofino [41]

Assuming the area below the line y=0 (i.e. x>1) does NOT count, the area to be rotated is shown in the graph attached.

A. Again, using Pappus's theorem,
Area, A = (2/3)*1*(1-(-1))=4/3 (2/3 of the enclosing rectangle, or you can integrate)
Distance of centroid from axis of rotation, R = (2-0) = 2
Volume = 2 π RA = 2 π 2 * 4/3 = 16 π / 3 (approximately = 16.76 units)

B. By integration, using the washer method
Volume = $2\pi\int_{-1}^1(1-x^2)(2-x)dx$
$=2\pi\int_{-1}^1(x^3-2x^2-x+2)dx$
$=2\pi[x^4/4-2x^3/3-x^2/2+2x]_{-1}^{1}$
$=2\pi([1/4-2/3-1/2+2]-[1/4+2/3-1/2-2])$
$=2\pi(8/3)$
= 16 π /3 as before

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3 years ago

In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

Oksi-84 [34.3K]

Answer:

a) $\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

b) From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

$P(Z\geq2.070)=1-P(Z$

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$

And we can use the z score defined as:

$z=\frac{\bar x -\mu}{\sigma_{\bar x}}$

And using this we got:

$P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

And using a calculator, excel or the normal standard table we have that:

$P(Z\geq2.070)=1-P(Z$

8 0

3 years ago

HELP ASAP TEST IS OMOST OVER!!!!!!!!!!

svet-max [94.6K]

Answer: C)46 ft

Step-by-step explanation:

We know that the circumference of a circle can be calculated with this formula:

$C=2\pi r$

Where "r" is the radius of the circle.

Since John is putting a fence around his garden that is shaped like a half circle and a rectangle, then we can find how much fencing he needs by making this addition:

$Fencing=\frac{2\pi r}{2}+2l+w$