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tankabanditka [31]

3 years ago

7

One angle of an isosceles triangle measures 110°. Which other angles could be in that isosceles triangle? Choose all that apply.

55 35 110 75

1 answer:

BigorU [14]3 years ago

8 0

Answer:

75 an 35 YEE YEE.

Step-by-step explanation:

im pretty sure

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All of the middle school mathletes solve math problems at the same pace. Working together, six mathletes complete all of the sum

guapka [62]

Answer:

5 1/2 hours

Step-by-step explanation:

by a factor of 5 means times 4

6*4=24

22/4=5.5 hours

8 0

3 years ago

Also need help on this

Ira Lisetskai [31]

Answer is 0.5
-2 divided by -4 equals 0.5
and same thing for the rest of the equation.

8 0

3 years ago

Read 2 more answers

Belle and Mabel are both 8-week-old puppies of different breeds. Belle has a mass of 12

balandron [24]

Answer:

3:1

Step-by-step explanation:

Take the fraction and simplify it:

12/4 = 3/1

6 0

3 years ago

The radius of a circle is (7x+3)cm. Write an expression to represent the area of the circle in simplified form

zepelin [54]

Answer:

A = π(7x + 3)² cm²

Step-by-step explanation:

A = πr² is the appropriate equation. If r = 7x + 3 cm, then the area of this particular circle is:

A = π(7x + 3)² cm²

3 0

2 years ago

A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are

Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

$P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764$

There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

(b) What is the probability that all three of the bulbs have the same rating?

$P = P_{1} + P_{2} + P_{3}$

$P_{1}$ is the probability that all three of them are 13-watt. So:

$P_{1} = \frac{8}{24}*\frac{7}{23}*\frac{6}{22} = 0.0277$

$P_{2}$ is the probability that all three of them are 18-watt. So:

$P_{2} = \frac{9}{24}*\frac{8}{23}*\frac{7}{22} = 0.0415$

$P_{3}$ is the probability that all three of them are 23-watt. So:

$P_{3} = \frac{7}{24}*\frac{6}{23}*\frac{5}{22} = 0.0173$

$P = P_{1} + P_{2} + P_{3} = 0.0277 + 0.0415 + 0.0173 = 0.0865$

There is a 8.65% probability that all three of the bulbs have the same rating.

(c) What is the probability that one bulb of each type is selected?

We have to permutate, permutation of 3(bulbs), with (1,1,1) repetitions(one for each type). So

$P = p^{3}_{1,1,1}*\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 3**\frac{8}{24}*\frac{9}{23}*\frac{7}{22} = 0.1245$

There is a 12.45% probability that one bulb of each type is selected.

3 0

2 years ago