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Which number is a common factor of 32, 48. and 80? 12 10 8 9

Alenkasestr [34]

Answer:

The answer is 8.

5 0

3 years ago

Jason cycled for 3 hours. He cycled 7 1/2 miles each hour for the first two hours and the distance he cycled for the third hour

suter [353]

Answer:

Total distance Jason cycled = 20 miles

Step-by-step explanation:

Distance covered in first 2 hours = 2( 7.5)

=15

Distance covered in the 3rd hour = 15/3

= 5

Dividing by 3 because the distance covered in the 3rd hour is 1/4 of the distance covered in first two hours. By dividing by 3 we will get 1/4 of the distance

Total Distance = Distance covered in first two hours

+ Distance covered in the 3rd hour

=15 + 5

Total Distance = 20

3 0

3 years ago

In a class of 18 students, 5 are math majors. A group of four students is chosen at random. (Round your answers to four decimal

ss7ja [257]

Answer:

(a) 0.2721

(b) 0.7279

(c) 0.2415

Step-by-step explanation:

(a) If we choose only one student, the probability of being a math major is $\frac{5}{18}$ (because there are 5 math majors in a class of 18 students). So, the probability of not being a math major is $\frac{18}{18} - \frac{5}{18} = \frac{13}{18}$ (we subtract the math majors of the total of students).

But there are 4 students in the group and we need them all to be not math majors. The probability for each one of not being a math major is $\frac{13}{18}$ and we have to multiply them because it happens all at the same time.

P (no math majors in the group) = $\frac{13}{18} *\frac{13}{18}*\frac{13}{18}*\frac{13}{18} = (\frac{13}{18}) ^4$ = 0.2721

(b) If the group has at least one math major, it has one, two, three or four. That's the complement (exactly the opposite) of having no math majors in the group. That means 1 = P (at least one math major) + P (no math major). We calculated this last probability in (a).

So, P (at least one math major) = 1 - P(no math major) = 1 - 0.2721 = 0.7279

(c) In the group of 4, we need exactly 2 math majors and 2 not math majors. As we saw in (a), the probability of having a math major in the group is 5/18 and having a not math major is $\frac{13}{18}$ . We need two of both, that's $\frac{5}{18}*\frac{5}{18}*\frac{13}{18}*\frac{13}{18}$ . But we also need to multiply this by the combinations of getting 2 of 4, that is given by the binomial coefficient $\binom{4}{2}$ .

So, P (exactly 2 math majors) = $\binom{4}{2}*(\frac{5}{18} )^2*(\frac{13}{18})^2$ = $\frac{4!}{2!2!}*\frac{25}{324}*\frac{169}{324}$ = 0.2415

7 0

4 years ago

5 quarts =how many fluid ounces

pishuonlain [190]

160 fluid ounces :) hope this helped

5 0

3 years ago

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10. A movie theater advertises that a family of two adults, one student, and one child between the ages of 3

BlackZzzverrR [31]

Answer:

Students ticket = $4

Child ticket = $1

Adult ticket = $5

Step-by-step explanation:

Let the price of the student ticket be $x and the price of a child ticket be $y. An adult ticket costs as much as the combined cost of a student ticket and a child ticket, so the price of one adult ticket is $(x+y).

1. A movie theater advertises that a family of two adults, one student, and one child between the ages of 3 and 8 can attend a movie for $15. Then

$2(x+y)+x+y=15$

2. If you purchase 1 adult ticket, 4 student tickets, and 2 child tickets for $23, then

$(x+y)+4x+2y=23$

Now, solve the system of two equations:

$\left\{\begin{array}{l}2(x+y)+x+y=15\\ \\(x+y)+4x+2y=23\end{array}\right.\Rightarrow \left\{\begin{array}{l}3(x+y)=15\\ \\5x+3y=23\end{array}\right.\\ \\\left\{\begin{array}{l}x+y=5\\ \\5x+3y=23\end{array}\right.\\ \\\left\{\begin{array}{l}y=5-x\\ \\5x+3(5-x)=23\end{array}\right.$

Solve the last equation

$5x+15-3x=23\\ \\2x=23-15\\ \\2x=8\\ \\x=4\\ \\y=5-x=5-4=1$

Students ticket = $4

Child ticket = $1

Adult ticket = $5

8 0

3 years ago

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