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3 years ago

Please help me with this D:

Mathematics

1 answer:

I am Lyosha [343]3 years ago

8 0

Yes, because $120 - 36 = (120\times 36) - 4236 = 84$

Hope that helped :D

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(x - 5)(x + 2) Polynomial

olya-2409 [2.1K]

X^2-3x-10. You solve it by foiling. I attached a photo explaining how to do it.

7 0

3 years ago

A cone and a triangular pyramid have a height of 9.3 m

Otrada [13]

Answer:

$x=17.1\ in$

Step-by-step explanation:

<u><em>The complete question is</em></u>

A cone and a triangular pyramid have a height of 9.3 m and their cross-sectional areas are equal at every level parallel to their respective bases. The radius of the base of the cone is 3 in and the other leg (not x) of the triangle base of the triangular pyramid is 3.3 in

What is the height, x, of the triangle base of the pyramid? Round to the nearest tenth

The picture of the question in the attached figure

we know that

If their cross-sectional areas are equal at every level parallel to their respective bases and the height is the same, then their volumes are equal

Equate the volume of the cone and the volume of the triangular pyramid

$\frac{1}{3}\pi r^{2}H=\frac{1}{3}[\frac{1}{2}(b)(h)H]$

simplify

$\pi r^{2}=\frac{1}{2}(b)(h)$

we have

$r=3\ in\\b=3.3\ in\\h=x\ in\\pi=3.14$

substitute the given values

$(3.14)(3)^{2}=\frac{1}{2}(3.3)(x)$

solve for x

$28.26=\frac{1}{2}(3.3)(x)$

$x=28.26(2)/3.3\\x=17.1\ in$

7 0

3 years ago

Read 2 more answers

A game is played with a spinner on a circle, like the minute hand on a clock. The circle is marked evenly from 0 to 100, so, for

zheka24 [161]

Answer:

The probability is 1/2

Step-by-step explanation:

The time a person is given corresponds to a uniform distribution with values between 0 and 100. The mean of this distribution is 0+100/2 = 50 and the variance is (100-0)²/12 = 833.3.

When we take 100 players we are taking 100 independent samples from this same random variable. The mean sample, lets call it X, has equal mean but the variance is equal to the variance divided by the length of the sample, hence it is 833.3/100 = 8.333.

As a consecuence of the Central Limit Theorem, the mean sample (taken from independant identically distributed random variables) has distribution Normal with parameters μ = 50, σ= 8.333. We take the standarization of X, calling it W, whose distribution is Normal Standard, in other words

$W = \frac{X - \mu}{\sigma} = \frac{X - 50}{8.333} \simeq N(0,1)$

The values of the cummulative distribution of the Standard Normal distribution, lets denote it $\phi$ , are tabulated and they can be found in the attached file, We want to know when X is above 50, we can solve that by using the standarization