A copy machine makes 231 copies in 5 minutes and 15 seconds. How many copies does it make per minute?

Mathematics

1 answer:

kotykmax [81]3 years ago

7 0

Put 5 min 15 sec into seconds (315 seconds) and set up a ratio to find how many copies are made every 60 seconds.

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A basketball court is being painted. The entire section below will be painted blue. Determine how many square feet of paint will

earnstyle [38]

Answer:

145.12 ft squared will be painted blue.
So the answer is B.

Explanation:

The area of the rectangle is 120 ft2 because 8 ft x 15 ft = 120 ft2

The area of the semi-circle is
because 3.14 x 4 (the radius) squared (or 16) is 50.24, then you divide it by 2 since it’s a semi-circle. 50.24 divided by 2 is 25.12.

Then you add the areas:

120 ft squared + 25.12 ft squared = 145.12 ft squared

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3 years ago

HELP ME PLEASE

Ghella [55]

Answer:

B) No, supplementary angles are either of two angles whose sum is 180 degrees.

C) Yes, the triangles are congruent because the triangles have exactly the same size and shape

Step-by-step explanation:

B) From the image below an 180 degree angle would be obtuse in which neither angle appears to be so they are not supplementary.

C) They are congruent because they are taking angles of the same angles as one another just on different sides of the shape.

4 0

2 years ago

Read 2 more answers

A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d

tankabanditka [31]

The selection of r objects out of n is done in

$C(n, r)= \frac{n!}{r!(n-r)!}$ many ways.

The total number of selections 10 that we can make from 6+7=13 students is

$C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}= 286$
thus, the sample space of the experiment is 286

A.
"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.

 $C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35$

P(all 6 girls chosen)=35/286=0.12

B.
"The probability that a randomly chosen team has 3 girls and 7 boys."

with the same logic as in A, the number of groups were all 7 boys are in, is

 $C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20$

so the probability is 20/286=0.07

C.
"The probability that a randomly chosen team has either 4 or 6 boys."

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is 0.12.

case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.

 $C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105$

the probability is 105/286=0.367

since case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"The probability that a randomly chosen team has 5 girls and 5 boys."

selecting 5 boys and 5 girls can be done in

 $C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126$

many ways,

so the probability is 126/286=0.44

6 0

3 years ago

Read 2 more answers

The numbers of blocked intrusion attempts on each day during the first two weeks of

Marrrta [24]

(a) Construct a 95% confidence interval for the difference between the average number of intrusion attempts per day before and after the change of firewall setting (assume equal variance)