Ask question

Ask question

All categories

3 years ago

6

Greg has 4 shirts: a white one, a black one, a red one, and a blue one. He also has two pairs of pants, one blue and one tan. Wh

at is the probability, if Greg gets dressed in the dark, that he winds up wearing the white shirt and tan pants?

2 answers:

nika2105 [10]3 years ago

7 0

P(WT)=(1/4)(1/2)=1/8

Katen [24]3 years ago

7 0

1/8 he winds up wearing the white shirt and tan pants

You might be interested in

Brand X costs $1.65 for 11 ounces. Brand Y costs 2 cents more per ounce. What is the cost of 15 ounces of brand Y?

34kurt

Answer:

$2.25

Step-by-step explanation:

You can find the answer by setting up a simple equation.

Brand X is 1.65 for 11 ounces

we can find the cost of one ounce of Brand X by using division

1.65/11

.15

this means that for every ounce of Brand X, it costs 15 cents

Since brand Y costs two cents more per ounce that would make it be 17 cents per ounce

now all you need to do is multiply and solve.

.15 * 15 = $2.25

8 0

2 years ago

Read 2 more answers

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

lesya [120]

D. 6 steps forward and 6 steps backward. 6=6

7 0

3 years ago

A circle has a radius of 9 cm and a sector of the circle has an arc length of 9. 7cm

frutty [35]

Answer:

S = R θ where θ is in radians

If θ = 2 pi then S = 2 pi R the circumference of the circle

θ = S / R = 9.7 / 9 = 1.08 rad

Since 1 rad = 360 / (2 pi) = 57.3 deg

the angle included = 1.08 rad * 57.3 deg/rad = 61.9 deg

7 0

2 years ago

One day in January the temperature dropped steadily from 3 degrees to negative 3 degrees. What was the change in temperature

AveGali [126]

6 degrees. ddrrgtexc7ehdb

6 0

3 years ago