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3 years ago

7

ASAP Someone please help me! PLEASE, I BEG My grade is going down and i can only count on u guys!

2 answers:

kirza4 [7]3 years ago

8 0

Additive inverses combine to get 0. So, k is -1.4. The sum is 0. Hopefully that will help.

bazaltina [42]3 years ago

4 0

An additive inverse is a number and it's positive or negative equal away from zero. Both numbers added together must equal zero. So 8 and -8 are additive inverses because they equal zero. 10.5 and -10.5 are additive inverses.

For your problem above, 1.4 is the additive inverse of -1.4 because added together they equal zero.

Given #= 1.4
k= -1.4
sum= 0

Hope this helps! :) Good luck!!

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Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha

const2013 [10]

Answer:

(a) The value of <em>a</em> is 53.35.

(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

$f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}$

Here, 22 < X < 55.

(a)

Compute the value of <em>a</em> as follows:

$P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35$

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

$P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17$

Thus, the value of <em>a</em> is 38.17.

(c)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95$

Thus, the value of <em>a</em> is 26.95.

(d)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63$

Thus, the value of <em>a</em> is 25.63.

(e)

Compute the value of <em>a</em> as follows:

$P(1.83\leq X\leq a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06$

Thus, the value of <em>a</em> is 12.06.

7 0

3 years ago

Rewrite the fraction in simplest form. 62/70

Ostrovityanka [42]

The answer is 31/35
Hope it helps!

8 0

3 years ago

The center of a circle is at (2, -5) and it's radius is 12. What is the equation of the circle?

Svetach [21]

The equation for a circle is (x-h)^2 +(y-k)^2=r^2
You just have to plug in your coordinates h is the x and k is the y.
The r is radius squared.
so, (x-2)^2+(y-(-5)^2=144
You have to be careful of those tricky negative values. When you distribute the y-(-5) it turns into y+5.
So your answer is D.

8 0

3 years ago

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Please help me I’m so lost

Svetllana [295]

Answer:

759.88 square inches

Step-by-step explanation:

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3 years ago

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Alisiya [41]

Answer:nhnc

add it mc

Step-by-step explanation:vgnhc

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3 years ago