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Blababa [14]
3 years ago
13

A.) How many ways can you select 1 flavor of ice cream, 1 topping and 1 drink if there are 5 flavors of ice cream, 3 toppings an

d 4 drinks to choose from?
B.) How many ways can you select a 4-digit passcode without repeating any digit?
C.) How many ways can you select 3 toppings for a pizza with 8 toppings to choose from?
D.) What is the probability of rolling a single die and it landing on a prime number? (Remember, the number 1 is not considered prime.)
E.) How many items are in the sample space for rolling a die and spinning a spinner with 4 different colored sections (red, yellow, green and blue)?
F.) What is the probability of rolling the die and spinning the 4-sectioned spinner described in part E, and getting the number 5 and the color red?
G.) If you wanted to purchase a size T-shirt that "most" people could wear, would you purchase the mean size, the median size or the mode size?
Mathematics
1 answer:
jeka943 years ago
7 0
A) you can combine any of the 5 ice-creams with any of the 3 toppings and any of the 4 drinks.

This means a total of 5*3*4=60 ways.

B) assuming that the first digit can be 0, we have 10 choices for the 1. digit, 9 for the 2., 8 for the third and 7 choices for the fourth digit.

In total, there are 10*9*8*7=5040 passwords we can create.

C) Selection of 3 objects out of 8, can be done in

C(8, 3)=8!/(3!5!)=(8*7*6)/3!=8*7=56 many ways.

D.

P(prime)=n(prime)/n{1, 2, 3, 4, 5, 6}=n{2, 3, 5}/6=3/6=1/2=0.5

E.

any of the 6 outcomes of rolling a die, can be combined with any of the 4 colors.

for example (1, red), (1, yellow), (1, green), (1, blue).

so in total, there are 6*4=24 items in the sample space.

F.
(5, red) is just one of the 24 items in the sample space, so its probability is 1/24= 0.0417

G.

the mode, as it is the size which appears the most.


Answers:

A.60

B.5040

C.56

D. 0.5

E.24

F.1/24= 0.0417

G.mode




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A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11
mixer [17]
Let A(t) denote the amount of salt in the tank at time t. We're given that the tank initially holds A(0)=100 lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}
\implies A'(t)+\dfrac{11}{200+33t}A(t)=484

Find the integrating factor:

\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}

Distribute \mu(t) along both sides of the ODE:

(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}
\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}
A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt
A(t)=22(200+33t)^{2/3}+C

Since A(0)=100, we get

100=22(200)^{2/3}+C\implies C\approx-652.39

so that the particular solution for A(t) is

A(t)=22(200+33t)^{2/3}-652.39

The tank becomes full when the volume of solution in the tank at time t is the same as the total volume of the tank:

200+(44-11)t=500\implies 33t=300\implies t\approx9.09

at which point the amount of salt in the solution would be

A(9.09)\approx733.47\text{ lb}
4 0
3 years ago
The Frostburg-Truth bus travels on a straight road from Frostburg Mall to Sojourner Truth Park. The mall is 5 miles west and 3 m
Mamont248 [21]

Answer:

D. 11.4 miles.

Step-by-step explanation:

Distance between the Center and the Mall = √(5^2 + 3^2) = √34.

Distance between The Park and the Center = √(4^2 + 4^2) = √32.

So the Distance between The Mall and the Park is √34 + √32

=  11.4 miles (answer).

3 0
3 years ago
40 points and brainliest to the best correct answer, questions in photo.
vodomira [7]

Answer:

2. w = -3

4. c = -2

6. r = 7

8. n = 4

Step-by-step explanation:

It's hard to read the photo, but I'll try.

2.

2^{w + 4} \times 2^{4w + 6} = 2^{2w + 1}

On the left side, you have a product of 2 powers with the same base.

Give the same base and add the exponents, as in the rule

a^m \times a^n = a^{m + n}

You get on the left side 2 to the sum of the exponents.

2^{w + 4 + 4w + 6} = 2^{2w + 1}

Simplify the long exponent on the left side.

2^{5w + 10} = 2^{2w + 1}

Now we have another rule we can use. If two powers are equal, and their bases are equal, then the exponents must be equal.

Here, 2 to the power 5w + 10 equals 2 to the power 2w + 1. Since both bases are 2 and are equal, then the exponents must be equal.

5w + 10 = 2w + 1

Subtract 2w from both sides. Subtract 10 from both sides.

3w = -9

Divide both sides by 3.

w = -3

4.

\dfrac{1}{5} = 5^{2c + 3}

We need to write the left side as a power of 5. Then we equate the exponents like we did in problem 2.

Recall the rule:

a^{-n} = \dfrac{1}{n}

Apply this rule to the left side in reverse.

5^{-1} = 5^{2c + 3}

Now we have two powers that are equal, both having the same base, 5, so the exponents must be equal.

2c + 3 = -1

2c = -4

c = -2

6.

216 = 6^{2r - 11}

This is the same idea as problem 4. Write the left side as a power of 6.

6^3 = 6^{2r - 11}

2r - 11 = 3

2r = 14

r = 7

r = 7

8.

4^{n} \times 4^{2n - 9} = 64

This problem is similar to problem 2.

4^{n + 2n - 9} = 4^3

4^{3n - 9} = 4^3

3n - 9 = 3

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n = 4

4 0
2 years ago
Edwin has $50.00 in his bank. He starts spending $2.50 each week on snacks at lunch
lukranit [14]

Answer:

um whats the question? if you want the equation for it it's 50 - 2.50x = y

Step-by-step explanation:

3 0
3 years ago
Which possible buffet prices could Noah could charge and still maintain the restaurant owner’s revenue requirements?
Eduardwww [97]

Answer:

$13 and $15

Step-by-step explanation:

correct prices

4 0
3 years ago
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