NEED HELP!!!!!EMERGENCY

Mathematics

1 answer:

Basile [38]3 years ago

8 0

Answer:

49.5 square inches

Step-by-step explanation:

We can split the figure into two easier shapes to find the area of; a triangle and a square.

The formula for a triangle is (b*h)1/2. Once we solve we get 24.5 or 24 1/2.

The formula for a square is l*w. Once we solve we get 25.

To find the area of the shaded region, we would add both area and get 49.5 or 49 1/2 square inches.

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Ask Your Teacher Consider a binomial random variable with n = 5 and p = 0.8. Let x be the number of successes in the sample. Eva

Readme [11.4K]

Answer:

0.942 is the required probability.

Step-by-step explanation:

We are given the following in the question:

x is a binomial random variable with n = 5 and p = 0.8.

Then,

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

We have to evaluate:

$P(x \geq 3) = P(x = 3) + P(x = 4) +P(x=5)\\\\= \binom{5}{3}(0.8)^3(1-0.8)^2 +\binom{5}{4}(0.8)^4(1-0.8)^1 +\binom{5}{5}(0.8)^5(1-0.8)^0\\\\= 0.2048 +0.4096+0.3276=0.942$

0.942 is the required probability.

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3 years ago

Aaliyah practices clarinet a total of 11.25 hours in d days. She practices the same amount of time each day. How many hours does

amm1812

6 hours per day lol

6 0

3 years ago

99 POINT QUESTION, PLUS BRAINLIEST!!!

sattari [20]

We know, that the <span>area of the surface generated by revolving the curve y about the x-axis is given by:

$\boxed{A=2\pi\cdot\int\limits_a^by\sqrt{1+\left(y'\right)^2}\, dx}$

In this case a = 0, b = 15, $y=\dfrac{x^3}{15}$ and:

$y'=\left(\dfrac{x^3}{15}\right)'=\dfrac{3x^2}{15}=\boxed{\dfrac{x^2}{5}}$

So there will be:

$A=2\pi\cdot\int\limits_0^{15}\dfrac{x^3}{15}\cdot\sqrt{1+\left(\dfrac{x^2}{5}\right)^2}\, dx=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\left(\star\right)\\\\-------------------------------\\\\
\int x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\,dx=\int\sqrt{1+\dfrac{x^4}{25}}\cdot x^3\,dx=\left|\begin{array}{c}t=1+\dfrac{x^4}{25}\\\\dt=\dfrac{4x^3}{25}\,dx\\\\\dfrac{25}{4}\,dt=x^3\,dx\end{array}\right|=\\\\\\$

$=\int\sqrt{t}\cdot\dfrac{25}{4}\,dt=\dfrac{25}{4}\int\sqrt{t}\,dt=\dfrac{25}{4}\int t^\frac{1}{2}\,dt=\dfrac{25}{4}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}= \dfrac{25}{4}\cdot\dfrac{t^{\frac{3}{2}}}{\frac{3}{2}}=\\\\\\=\dfrac{25\cdot2}{4\cdot3}\,t^\frac{3}{2}=\boxed{\dfrac{25}{6}\,\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}}\\\\-------------------------------\\\\$

$\left(\star\right)=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\dfrac{2\pi}{15}\cdot\dfrac{25}{6}\cdot\left[\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}\right]_0^{15}=\\\\\\=
\dfrac{5\pi}{9}\left[\left(1+\dfrac{15^4}{25}\right)^\frac{3}{2}-\left(1+\dfrac{0^4}{25}\right)^\frac{3}{2}\right]=\dfrac{5\pi}{9}\left[2026^\frac{3}{2}-1^\frac{3}{2}\right]=\\\\\\=
\boxed{\dfrac{5\Big(2026^\frac{3}{2}-1\Big)}{9}\pi}$

Answer C.
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3 0

3 years ago

I WILL GIVE BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

NemiM [27]

Answer:

$- (x - 1) {}^{2} + 5$