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3 years ago

((60 points to whoever answers all of these))

Mathematics

2 answers:

scZoUnD [109]3 years ago

6 0

1 9 squared
2 15 cubed
3 14 cubed
4 11 to the power of 5
5 16 to the power of 4

Readme [11.4K]3 years ago

3 0

1. 9^2

2. 15^3

3. 14^3

4. 11^5

5. 16^4

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Donovan took a math test and got 20 correct questions and 5 incorrect answers. What was the percentage of correct answers?

solong [7]

ANSWER

80%

EXPLANATION

Donovan got 20 questions correct and 5 questions incorrect.

This means that the total number of questions he attempted in the test is the sum of correct and incorrect questions:

Total = 20 + 5

Total = 25

To find the percentage of correct answers, we have to divide the number of correct answers by the total number of questions attempted and multiply by 100 (per cent).

That is:

$\begin{gathered} \frac{20}{25}\cdot\text{ 100 = }\frac{4}{5}\cdot\text{ 100} \\ =\text{ 80\%} \end{gathered}$

That is the percentage of correct answers.

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1 year ago

What is the value of x?

Marrrta [24]

Answer:

X is 39

Step-by-step explanation:

Explanation in picture above

4 0

3 years ago

Help me with this please

o-na [289]

Answer:

C is correct

Step-by-step explanation:

3+6=9

2 sides on a triangle must be greater than the other side

6 0

3 years ago

A bag contains 6 red apples and 5 yellow apples. 3 apples are selected at random. Find the probability of selecting 1 red apple

tester [92]

Answer: The required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

Step-by-step explanation: We are given that a bag contains 6 red apples and 5 yellow apples out of which 3 apples are selected at random.

We are to find the probability of selecting 1 red apple and 2 yellow apples.

Let S denote the sample space for selecting 3 apples from the bag and let A denote the event of selecting 1 red apple and 2 yellow apples.

Then, we have

$n(S)=^{6+5}C_3=^{11}C_3=\dfrac{11!}{3!(11-3)!}=\dfrac{11\times10\times9\times8!}{3\times2\times1\times8!}=165,\\\\\\n(A)\\\\\\=^6C_1\times^5C_2\\\\\\=\dfrac{6!}{1!(6-1)!}\times\dfrac{5!}{2!(5-2)!}\\\\\\=\dfrac{6\times5!}{1\times5!}\times\dfrac{5\times4\times3!}{2\times1\times3!}\\\\\\=6\times5\times2\\\\=60.$

Therefore, the probability of event A is given by

$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{60}{165}=\dfrac{4}{11}\times100\%=36.36\%.$

Thus, the required probability of selecting 1 red apple and 2 yellow apples is 36.36%.

4 0

3 years ago

Please answer this correctly without making mistakes

SCORPION-xisa [38]

First, multiply 1,127 & 1 by 2
1127 x 2 = 2254
1 x 2 = 2

so that’s 2,254 cups & 2 fl. ounces
since 8 fl. oz. make a cup, and we only have 2, the answer stays as it is

2,254 cups & 2 fl. ounces

4 0

3 years ago