All categories

3 years ago

Pls add me on ins if you gonna help me with my homework

Mathematics

1 answer:

Darina [25.2K]3 years ago

4 0

Answer:

The image does not show up

Step-by-step explanation:

You might be interested in

Hello all!

jasenka [17]

Answer:

sat: 28:63

sun:32:72

Step-by-step explanation:

thats is.

6 0

2 years ago

Read 2 more answers

You work 40 hours per week and spend 5 hours per week commuting to and from work. Your friend works 56 hours per week and spends

Burka [1]

Answer:

the answer is yes

Step-by-step explanation:

your ratio..40:5

friends ratio..56:7

each quantity is 8

3 0

3 years ago

A multiple-choice examination has 15 questions, each with five answers, only one of which is correct. Suppose that one of the st

Alex

Answer:

0.0111% probability that he answers at least 10 questions correctly

Step-by-step explanation:

For each question, there are only two outcomes. Either it is answered correctly, or it is not. The probability of a question being answered correctly is independent from other questions. So we use the binomial probability distribution 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.

A multiple-choice examination has 15 questions, each with five answers, only one of which is correct.

This means that $n = 15, p = \frac{1}{5} = 0.2$

What is the probability that he answers at least 10 questions correctly?

$P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

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

$P(X = 10) = C_{15,10}.(0.2)^{10}.(0.8)^{5} = 0.0001$

$P(X = 11) = C_{15,11}.(0.2)^{11}.(0.8)^{4} = 0.000011$

$P(X = 12) = C_{15,12}.(0.2)^{12}.(0.8)^{3} \cong 0$

$P(X = 13) = C_{15,13}.(0.2)^{13}.(0.8)^{2} \cong 0$

$P(X = 14) = C_{15,14}.(0.2)^{14}.(0.8)^{1} \cong 0$

$P(X = 15) = C_{15,15}.(0.2)^{15}.(0.8)^{0} \cong 0$

$P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.0001 + 0.000011 = 0.000111$

0.0111% probability that he answers at least 10 questions correctly

3 0

2 years ago

How to solve 4x-c=k solve for x

Tresset [83]

Answer:

$x = \frac{k + c}{4}$

Step-by-step explanation:

Given:

$4x - c = k$

Solve for:

$x$

Solution:

(1) Let's move all of components that do not contain $x$ (our target) to the right side (notice the change of sign of $c$ from negative to positive):

$4x = k + c$

(2) Let's divide both sides of equation by 4:

$\frac{4x}{4} = \frac{k + c}{4}$

(3) Now, we simplify the left side:

$x = \frac{k + c}{4}$

This is the solution.

Hope this helps!

3 0

3 years ago

Hi everyone can someone help me out with this question please

luda_lava [24]

A) After any given year the amount of money would be: [(1000*.04)x]+1000
and x stands for the number of years.
B) The interest is 4%.
C) There would be $3,480.00 after 4 years.

5 0

2 years ago