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3 years ago

Explain how you can use the picture to show 391 ÷ 23 = 17

Mathematics

1 answer:

Pavlova-9 [17]3 years ago

5 0

Answer:so the top of the rectangle is the answer so 10 +7

Step-by-step explanation:

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Find the quotient. 0.42 -0.7 HELP PLEASE

Simora [160]

Answer:

-0.28

Step-by-step explanation:

6 0

3 years ago

Given that a 90% confidence interval for the mean height of all adult males in Idaho measured in inches was [62.532, 76.478]. Wh

NNADVOKAT [17]

Answer:

The point estimate used to estimate the mean height of all adult males in Idaho is 69.505 inches.

Step-by-step explanation:

Each confidence interval has two bounds, the lower bound and the upper bound. The points estimate used to estimate the mean is the halfway point between those two bounds, that is, the sum of those two bounds divided by two.

In this problem, we have that:

Lower bound: 62.532

Upper bound: 76.478

Point estimate: (62.532 + 76.478)/2 = 69.505

The point estimate used to estimate the mean height of all adult males in Idaho is 69.505 inches.

3 0

3 years ago

The number name of 70080067 In Indian System

zloy xaker [14]

Answer:

Seven crore eighty thousands sixty-seven.

Step-by-step explanation:

70080067

Seven crore eighty thousands sixty-seven.

3 0

3 years ago

Use spherical coordinates. evaluate (9 − x2 − y2) dv, where h is the solid hemisphere x2 + y2 + z2 ≤ 4, z ≥ 0.

avanturin [10]

In spherical coordinates, we have

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}$

which gives volume element

$\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

and so the triple integral is given by

$\displaystyle\iiint_H(9-x^2-y^2)\,\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/2}\int_{\rho=0}^{\rho=2}(9-\rho^2\sin^2\varphi)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$
$=\displaystyle2\pi\int_{\varphi=0}^{\varphi=\pi/2}\int_{\rho=0}^{\rho=2}(9\rho^2\sin\varphi-\rho^4\sin^3\varphi)\,\mathrm d\rho\,\mathrm d\varphi$
$=\dfrac{592\pi}{15}$

4 0

3 years ago

CAN SOMEONE PLEASE HELP ME ASAP!!

Helga [31]

Answer:

Step-by-step explanation:

i dont really know man im sorry

7 0

3 years ago