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3 years ago

Plz answer quickly will mark brainless

Mathematics

2 answers:

bezimeni [28]3 years ago

7 0

Answer:

<h2>She must read more than 5 more hours</h2><h2>D.</h2>

12x5=60+10=70 pages.

She reads 12 pages per hour, so if she read for 5 hours, she would have read 60 pages. She has already read 10 pages, so that would equal 70 Pages.

den301095 [7]3 years ago

5 0

Answer:

She must read more than 5 hours

Step-by-step explanation:

equation 12h+10>70

12h>60

h>5

So she must read more than 5 hours

(can I get brainliest now)

You might be interested in

Suppose users share a 3 Mbps link. Also suppose each user requires 150 kbps when transmitting, but each user transmits only 10 p

djverab [1.8K]

Answer:

The probability that a given user is transmitting is 0.1 = 10%.

The probability that at any given time, exactly n users are transmitting simultaneously is $P(X = n) = C_{120,n}.(0.1)^{n}.(0.9)^{120-n}$ .

0.0048 = 0.48% probability that there are 21 or more users transmitting simultaneously.

Step-by-step explanation:

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.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Each user transmits only 10 percent of the time.

This means that $p = 0.1$

Find the probability that a given user is transmitting.

The probability that a given user is transmitting is 0.1 = 10%.

Suppose there are 120 users.

This means that $n = 120$

Find the probability that at any given time, exactly n users are transmitting simultaneously.

This is $P(X = n)$ .

This n is different from the n of total number of users(120 in this case) from the standard binomial formula. This is the number of successes, which is the equivalent of x. So

$P(X = n) = C_{120,n}.(0.1)^{n}.(0.9)^{120-n}$

The probability that at any given time, exactly n users are transmitting simultaneously is $P(X = n) = C_{120,n}.(0.1)^{n}.(0.9)^{120-n}$

Find the probability that there are 21 or more users transmitting simultaneously.

Now we use the binomial approximation to the normal. We have that:

$\mu = E(X) = np = 120*0.1 = 12$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{120*0.1*0.9} = 3.2863$

Using continuity correction, this probability is $P(X \geq 21 - 0.5) = P(X \geq 20.5)$ , which is 1 subtracted by the pvalue of Z when X = 20.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20.5 - 12}{3.2863}$

$Z = 2.59$

$Z = 2.59$ has a pvalue of 0.9952

1 - 0.9952 = 0.0048

0.0048 = 0.48% probability that there are 21 or more users transmitting simultaneously.

7 0

3 years ago

The perimeter of a rectangular book is given as 16.84 cm. Find the length of the rectangular book (in decimals) whose width is 1

Damm [24]

Answer:

5.8 is answer i hope i helped u

Step-by-step explanation:

3 0

3 years ago

How can we use a graph to tell that two populations are equal?

Dominik [7]

Answer: You can see if the graphs of the two populations meet at a given point.

Step-by-step explanation:

If we have a graph of population vs time

such that y = population and x = time, two different populations will be exactly the same for a given time if we have an intersection of the two graphs. The value (x0, y0) where the graphs cross is the time = x0 and the population y0, this means that both populations have y0 individuals at the same time.

4 0

3 years ago

Choose true or false:

Shalnov [3]

Answer:

False.

Step-by-step explanation:

y+9=0

y=-9

hope you have a good day/night! :)

6 0

2 years ago

Please help!!

Tpy6a [65]

Answer:

theoposite of 0is0..thisis acounter examplebecasue 0 is neither apositive or negative

Step-by-step explanation:

4 0

3 years ago