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3 years ago

In a parking lot with 250 parking spaces 40% how many of them are vacant and occupied

1 answer:

3 0

If that 40% is how many are occupied then- 100 are occupied 150 are vacant
if that 40% is how many are vacant then- 100 are vacant and 150 are occupied

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Simplify<br><img src="https://tex.z-dn.net/?f=4%282%20%2B%203i%29%20-2%20%286%20-%208i%29" id="TexFormula1" title="4(2 + 3i) -2

seraphim [82]

Hi there.

4(2 + 3i) - 2(6 - 8i); apply the distributive property.
8 + 12i - 12 - 16i; combine like-terms.
-4 - 4i is your final expression.

I hope this helps!

6 0

3 years ago

The solution to 2 x 27 will be what kind of number

SCORPION-xisa [38]

2*27= 54

54 is a rational number

I hope that's help :)

5 0

3 years ago

(a) A parachutist lands at a point on the line between the points A and B, and the target is an operation at A. The operation fa

mr Goodwill [35]

Answer:

a) $\frac{B-\frac{A+5B}{6}}{B-A}= \frac{6B -A-5B}{6(B-A)}=\frac{B-A}{6(B-A)}=\frac{1}{6}$

b) $P(X>1) = 1-P(X\leq 1)=1-\int_{0}^1 \frac{1^{2} x^{2-1} e^{- x}}{\gamma(2)}=0.736$

Step-by-step explanation:

Part a

We assume that the parachutist lands at random point in the interval (x=A,y=B) we have a continuous random variable X. And the distribution of X would be uniform $Y\sim Unif(A,B)$ . And the density function would be given by:

$f(x) =\frac{1}{B-A} , A$

And 0 for other case.

The operation fails, if parachutist's distance to A is more than five times as much as her distance to B.

So the point P in the interval (A,B) at which the distance to A is exactly 5 times the distance to B is given by:

$P= A + \frac{5}{6} (B-A)= \frac{6A +5B -5A}{6}=\frac{A+5B}{6}$

And we can find the probability desired like this:

$P(d(P,A) \geq 5 d(P,B))= P(\frac{A+5B}{6} < X< B)$

And from the cumulative distribution function of X ficen by $F(X)\frac{X-A}{B-A}$ we got:

$\frac{B-\frac{A+5B}{6}}{B-A}= \frac{6B -A-5B}{6(B-A)}=\frac{B-A}{6(B-A)}=\frac{1}{6}$

Part b

For this case we assume that $X\sim Gamma (2,1)$

On this case we assume that $\alpha=2, \beta= 1$

The density function for the Gamma distribution is given by:

$P(X)= \frac{\beta^{\alpha} x^{\alpha-1} e^{-\beta x}}{\gamma(\alpha)}$

And on this case we can find the probability using the complement rule like this:

$P(X>1) = 1-P(X\leq 1)=0.736$

We can solve this problem with the following excel code:

"=1-GAMMA.DIST(1;2;1;TRUE)"

And if we do it by hand we need to do this:

$P(X>1) = 1-P(X\leq 1)=1-\int_{0}^1 \frac{1^{2} x^{2-1} e^{- x}}{\gamma(2)}=0.736$

6 0

3 years ago

??? Please help vcyvhxrgjvuvrx

Llana [10]

Answer:

61, 61, 62, 64, 64, 66, 69, 69, 70, 72, 73, 74, 78, 78
69
64
73

<---------------------------------------------------------------------------------------------------------->

10 20 30 40 50 60 70 80 90 100

×--×-×-----×-----×

7 0

3 years ago

Help me please!

telo118 [61]

Answer:

$f^{-1}(6) = 50$

Step-by-step explanation:

Given

$f(x) = \sqrt{2x} - 4$

Required

Find $f^{-1}(6)$

First, we calculate the inverse function

$f(x) = \sqrt{2x} - 4$

Express f(x) as y

$y = \sqrt{2x} - 4$

Swap the positions of x and y

$x = \sqrt{2y} - 4$

Solve for y: Add 4 to both sides

$4 + x = \sqrt{2y} - 4+4$

$4 + x = \sqrt{2y}$

Square both sides

$(4 + x)^2 = 2y$

Divide both sides by 2

$y = \frac{(4 + x)^2}{2}$

Express y as an inverse function

$f^{-1}(x) = \frac{(4 + x)^2}{2}$

Next, solve for: $f^{-1}(6)$

Substitute 6 for x

$f^{-1}(6) = \frac{(4 + 6)^2}{2}$

$f^{-1}(6) = \frac{(10)^2}{2}$

$f^{-1}(6) = \frac{100}{2}$

$f^{-1}(6) = 50$

7 0

3 years ago