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3 years ago

9

Help me please I don’t know what to do what’s the answer

1 answer:

shtirl [24]3 years ago

7 0

YAY!!! ANOTHER PERSON FRON ED-GENUITY!!!
All of the letters are equal to 90 degrees so the answer would be C.
Hope I helped!

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Solve 3x1 = 12. Explain your reasoning.

Yuki888 [10]

The left side
3
3
does not equal to the right side
12
12
, which means that the given statement is false.
False

6 0

3 years ago

Read 2 more answers

5 books cost 20 work out the cost of 9 books

Eduardwww [97]

9 books cost 32

Step-by-step explanation:

5 books =20

1 book= 5÷20 = 4

9 books = 4×9 = 32

7 0

3 years ago

Read 2 more answers

I need help finding the answer. Please tell me what it is and how you got it

guapka [62]

rather simple, simply count how many zeros are there in each and that'll be the exponent of the base 10.

$\bf 1\stackrel{\leftarrow \qquad \textit{23 zeros}\qquad \to }{00,000,000,000,000,000,000,000}\implies \stackrel{Mars}{10^{23}~Kgs} \\\\\\ 1\stackrel{\leftarrow \qquad \textit{26 zeros}\qquad \to }{00,000,000,000,000,000,000,000,000}\implies \stackrel{Neptune}{10^{26}~Kgs}$

3 0

4 years ago

if you roll a fair 6-sided die 9 times, what is the probability that at least 2 of the rolls come up as a 3 or a 4?

Kay [80]

Using the binomial distribution, it is found that there is a 0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For each die, there are only two possible outcomes, either a 3 or a 4 is rolled, or it is not. The result of a roll is independent of any other roll, hence, the <em>binomial distribution</em> is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

There are 9 rolls, hence $n = 9$ .

Of the six sides, 2 are 3 or 4, hence $p = \frac{2}{6} = 0.3333$

The desired probability is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

$P(X < 2) = P(X = 0) + P(X = 1)$

Then

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{9,0}.(0.3333)^{0}.(0.6667)^{9} = 0.026$

$P(X = 1) = C_{9,1}.(0.3333)^{1}.(0.6667)^{8} = 0.117$

Then:

$P(X < 2) = P(X = 0) + P(X = 1) = 0.026 + 0.117 = 0.143$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.143 = 0.857$

0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For more on the binomial distribution, you can check brainly.com/question/24863377

7 0

3 years ago

Two balls are drawn at random from an urn containing six white and nine red balls. Recall the equatio n for an. r) given below.

storchak [24]

Answer: (a) $\dfrac{1}{7}$ (b) $\dfrac{12}{35}$

Step-by-step explanation:

Given: Number of white balls : 6

Number of red balls = 9

Total balls = 15

(a) The probability that both balls are white is given by :-

$\dfrac{^6C_2}{^{15}{C_2}}\\\\=\dfrac{\dfrac{6!}{2!(6-2)!}}{\dfrac{15!}{2!(15-2)!}}=\dfrac{1}{7}$

∴ The probability that both balls are white is $\dfrac{1}{7}$ .

(b) The probability that both balls are red is given by :-

$\dfrac{^9C_2}{^{15}C_2}\\\\=\dfrac{\dfrac{9!}{2!(9-2)!}}{\dfrac{15!}{2!(15-2)!}}=\dfrac{12}{35}$

∴ The probability that both balls are red is $\dfrac{12}{35}$ .

7 0

3 years ago