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3 years ago

3. Please help me with this question

Mathematics

1 answer:

bagirrra123 [75]3 years ago

6 0

Answer:

Step-by-step explanation:

Multiply the exponents together

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Help me plz! if you can

IgorLugansk [536]

Answer: 180/540
Simplified term would be 1/3
The answer is b

3 0

3 years ago

9 times a number increased by 2

sladkih [1.3K]

Answer:

9x + 2

Step-by-step explanation:

Because you are multiplying an unknown number by 9, it would be considered 9*x or 9x.

Then, you add 2, because when you increase in math, you are adding.

Therefore, the final product is 9x + 2.

5 0

3 years ago

Read 2 more answers

: Maria attended a Chicago Cubs baseball game last night. The ticket to get into the game was $39.00 and hot dogs cost $5.50 eac

PIT_PIT [208]

Answer: 3

Step-by-step explanation:

At first you would subtract 39 from 60. Then you would do 21/5.5 and you would get 3.818… and due to the fact you can't go over 60 you would round to 3 and not 4

8 0

3 years ago

Write the number in expanded form 36.13

BabaBlast [244]

There are two ways you can write it:
1. 30 + 6 + (1 × 1/10) + (3 × 1/100)
2. 30 + 6 + 0.1 + 0.03
Either is correct :)

4 0

3 years ago

Read 2 more answers

Question 4 of 10, Step 1 of 1 1/out of 10 Correct Certify Completion Icon Tries remaining:0 The Magazine Mass Marketing Company

erastovalidia [21]

Answer:

0.0105 = 1.05% probability that no more than 3 of the entry forms will include an order.

Step-by-step explanation:

For each entry form, there are only two possible outcomes. Either it includes an order, or it does not. The probability of an entry including an order is independent of any other entry, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

The Magazine Mass Marketing Company has received 16 entries in its latest sweepstakes.

This means that $n = 16$

They know that the probability of receiving a magazine subscription order with an entry form is 0.5.

This means that $p = 0.5$

What is the probability that no more than 3 of the entry forms will include an order?

At most 3 including an order, which is:

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

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

$P(X = 0) = C_{16,0}.(0.5)^{0}.(0.5)^{16} \approx 0$

$P(X = 1) = C_{16,1}.(0.5)^{1}.(0.5)^{15} = 0.0002$

$P(X = 2) = C_{16,2}.(0.5)^{2}.(0.5)^{14} = 0.0018$

$P(X = 3) = C_{16,3}.(0.5)^{3}.(0.5)^{13} = 0.0085$

Then

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0 + 0.0002 + 0.0018 + 0.0085 = 0.0105$

0.0105 = 1.05% probability that no more than 3 of the entry forms will include an order.

6 0

2 years ago