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3 years ago

On a test hailey answered 64 out of 75 questions correctly. What percent of her answers were correct? With the work please!!

Mathematics

2 answers:

Airida [17]3 years ago

8 0

Set up a proportion

64/75= x/100

Cross multiply:
64*100=6400
6400/75= 85.33333....=x

Final answer: 85.333...%

amm18123 years ago

6 0

85.333%
Because when you are looking for a percentage you have to simplify

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Type the correct answer in the box.

natita [175]

Answer:

250 cm^3.

Step-by-step explanation:

Volume of the prism

= 12 * 5^2

= 300 cm^3.

Volume of the pyramid

= 1/3 * 25 * 6

= 50 cm^3

The volume of the space

= 300 - 50

250 cm^3.

4 0

3 years ago

What is a to the -6th power using a positive exponent

pentagon [3]

$a^{-6}=\left(\dfrac{1}{a}\right)^6=\dfrac{1}{a^6}$

5 0

3 years ago

What is the value of x? Enter your answer in the box. x =

Schach [20]

Answer:

The answer will be like this

X =20

6 0

3 years ago

TV studio has brought in 9 boy kittens and 6 girl kittens for a cat food commercial. The director is going to choose 9 of these

Amanda [17]

Answer:

probability = 0.2517

Step-by-step explanation:

given data

boy kittens = 9

girl kittens = 6

choose kittens at random = 9

solution

total kitten are = 9 + 6 = 15

first we get here total no of probability that is

n(s) = 15 C 9

n(s) = $\frac{15!}{9!(15-9)!}$

n(s) =5005

and

total way to chose 5 boy kittens is = 9 C 3

n(3) = $\frac{9!}{3!(9-3)!}$

n(3) = 84

and

total way to chose 4 girl kittens is = 6 C 4

n(4) = $\frac{6!}{4!(6-4)!}$

n(4) = 15

total way to chose 5 boy kittens and 4 girl kittens is

total way = 84 ×15 = 1260

so probability that the director chooses 5 boy kittens and 4 girl kittens is

probability = $\frac{1260}{5005}$

probability = 0.2517

6 0

4 years ago

AaBbCc x AaBbCc (use for all 3 questions) 1. What is the probability that this individual will: AABBcc 2. What is the probabilit

tankabanditka [31]

Answer: 1. P = 1/64

2. P = 1/32

3. P = 1/8

Step-by-step explanation:

In genetics, an ofspring inherits one copy of an allele of each parent, which can be described in the tables below called Punnett Square:

For crossing Aa x Aa:

A a

A AA Aa

a Aa aa

For crossing Bb x Bb:

B b

B BB Bb

b Bb bb

For crossing Cc x Cc:

C c

C CC Cc

c Cc cc

We can separate them because they are assorted independently.

For offspring with genotype AABBcc, probability will be:

P(AA) = 1/4

P(BB) = 1/4

P(cc) = 1/4

As all three probabilities has to happen at the same time, it is a "E" rule:

P(AABBcc) = $(\frac{1}{4}) (\frac{1}{4}) (\frac{1}{4})$

P(AABBcc) = 1/64

Probability for the individual to be AABBcc is 1/64 or 1.56%.

Genotype AaBBcc:

P(Aa) = 2/4 = 1/2

P(BB) = 1/4

P(cc) = 1/4

P(AaBBcc) = $(\frac{1}{2}) (\frac{1}{4}) (\frac{1}{4})$

P(AaBBcc) = 1/32

Probability for the individual to be AaBBcc is 1/32 or 3.12%

Genotype AaBbCc:

P(Aa) = 1/2

P(Bb) = 1/2

P(Cc) = 1/2

P(AaBbCc) = $(\frac{1}{2}) (\frac{1}{2}) (\frac{1}{2})$

P(AaBbCc) = 1/8

Probability for the individual to be AaBbCc is 1/8 or 12.5%.

4 0

3 years ago