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anzhelika [568]
3 years ago
11

1 point

Mathematics
1 answer:
s2008m [1.1K]3 years ago
6 0

Answer:

i do not know

Step-by-step explanation:

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A cat had a litter of 5 kittens. If each sex is equally likely, what is the probability that 3 or less were female kittens?
ASHA 777 [7]

Answer:

0.8125

Step-by-step explanation:

In this question, we are tasked with calculating the probability that 3 or less of her kittens were female.

Since each bsex is of likely probability, the probability of a male kitten = probability of a female kitten = 0.5

Now to calculate for 3 or less female kitten we are calcualting P(f) ≤ 3

In each case, we use the Bernoulli approximation

P(f) ≤ 3 = 5C3 m^{2} f^{3} + 5C2 m^{2}  f^{3} + 5C1 m^{4}  f + 5C0 m^{5}  f^{0}

Where m is the probability of a male kitten and f is the probability of having a female kitten with both values = 0.5

P(f) ≤ 3 =(0.3125) + (0.3125) + (0.15625) + (0.03125) = 0.8125

8 0
3 years ago
Solve y=x²+7 for x for apexvs homeschool 2019​
Molodets [167]

Hello!

To solve for x means to isolate x:

y = x^2 + 7

y - 7 = x^2

sqrt(y - 7) = x

Answer:

x = sqrt(y - 7)

Hope this helps if it did please make brainliest!

7 0
3 years ago
sally bought a hot pink motorized scooter for $5 000. online , she read that .the scooter is reduced each year by about 18% how
saveliy_v [14]
You need to set it up like this: Initial amount ( 1 - Depreciation Rate) ^ Time in years. So: 5000(1-0.18)^4 = 5000(0.82)^4=2260.6088, rounded to the nearest cent is $2260.61.
7 0
3 years ago
Which of the following are solutions to the equation below?
nalin [4]

\qquad\qquad\huge\underline{{\sf Answer}}

Let's solve the given equation ~

\qquad \tt \dashrightarrow \:4 {x}^{2}  + 20x + 25 = 49

\qquad \tt \dashrightarrow \:4 {x}^{2}  + 20x + 25  -  49 = 0

\qquad \tt \dashrightarrow \:4 {x}^{2}  + 20x - 24 = 0

\qquad \tt \dashrightarrow \:4( {x}^{2}  + 5x - 6) = 0

\qquad \tt \dashrightarrow \: {x}^{2}  + 5x - 6 = 0

\qquad \tt \dashrightarrow \: {x}^{2}  + 6x - x - 6 = 0

\qquad \tt \dashrightarrow \:x(x + 6) - 1(x + 6) = 0

\qquad \tt \dashrightarrow \:(x + 6)(x - 1) = 0

Hence, we get -6 and 1 as our roots ~

So, the correct choices are : B and D

3 0
2 years ago
Find the zeros of each function by using a graph and a table. f(x)=x^2+2x–24.
Alborosie

Explanation

Step 1

f(x)=x^2+2x-24

Zeros

A(-6,0) B(4,0)

because

\begin{gathered} f(x)=x^2+2x-24 \\ f(-6)=(-6)^2+2(-6)-24=36-12-24=0 \\ f(4)=4^2+2\cdot4-24=16+8-24=0 \end{gathered}

Step 2

table

\begin{gathered} (-6,0) \\ (4,0) \\ f(-1)=-1^2+2\cdot-1-24=1-2-24=-25 \\ (1,-25) \\  \end{gathered}

I hope this helps you

5 0
1 year ago
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