What is the gcf of 36 &90

Figure a is dilated by a scale factor of 1/2 to form figure b. What is the length of x?

PIT_PIT [208]

Answer:

The correct option is 3. The value of x is 8 inches.

Step-by-step explanation:

It is given that Figure A is dilated by a scale factor of 1/2 to form Figure B. It means both figure are similar and corresponding sides of similar figures are proportional.

$\text{scale factor}=\frac{\text{Side of image}}{\text{Side of preimage}}$

$\frac{1}{2}=\frac{x}{16}$

Multiply both sides by 16.

$\frac{16}{2}=x$

$8=x$

Since scale factor is 1/2, so we can directly say that the side length of image is half of the corresponding side of preimage.

The length of x is 8 inches. Therefore the correct option is 3.

3 0

3 years ago

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Which equation represents a line parallel to the line shown on the graph?

Mama L [17]

Answer:

-3x+3

Step-by-step explanation:

The equation to the line shown is formed as follows.

It passes through the points (-6,0) and (-8,6)

The gradient of the line=Δy/Δx

=(y₂-y₁)/(x₂-x₁)

=(6-0)/(-8--6)

=6/-2

=-3

The line parallel to the line shown has the same gradient i.e -3

Therefore the line in question is

-3x+3.

7 0

3 years ago

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

8 0

2 years ago

A company hauls its 1 cubic foot boxes in the truck as shown.

Alenkinab [10]

Answer:

96

Step-by-step explanation:

3*4*8=96

4 0

2 years ago

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(5x - 1, if x < -2<br> Evaluate f(x) =<br> Lx+ 3, if x > -2<br> when z = 3.

Brilliant_brown [7]

Answer:

(5x - 1, if x < -2

Evaluate f(x) =

Lx+ 3, if x > -2

when z = 3.

Step-by-step explanation:

5 0

2 years ago