All categories

3 years ago

Help me ASAP for this question

Mathematics

2 answers:

erastovalidia [21]3 years ago

8 0

So it is 7 because in the question x =14 so it’s 14-7 which is 7

ludmilkaskok [199]3 years ago

7 0

Answer:

Your answer is 7

Hope this helps!

Step-by-step explanation:

You plug 14 in for x then just do 14-7 and you will get 7

You might be interested in

Pleaseee help me.

Alex777 [14]

Answer:

y = 4x + 2

4 is the slope and 2 is what y will be at x = 0.

7 0

3 years ago

Read 2 more answers

Help picture below problem 6

r-ruslan [8.4K]

Answer:

option B is the correct answer

Step-by-step explanation:

mark me brainliest

6 0

2 years ago

What does 10 x 5 equal?

andriy [413]

I think it should be 50, really sorry if it wasn't multiplying.

7 0

3 years ago

Read 2 more answers

A thermometer is taken from an inside room to the outside, where the air temperature is 20° f. after 1 minute the thermometer re

fomenos

The change in the temperature is proportional to the difference in temperatures. Let's call the initial temperature of the room and the thermometer $E_0$ and the current temperature $E(t)$ where $t$ is the elapsed time in minutes. The outside temperature is 20 F.

Newton's Law of Cooling:

$E(t) - 20 = (E_0 - 20) e^{- k t}$

We're given $E(1)=70 \textrm{ and } E(5)=45$

$70-20 = (E_0 - 20) e^{-k}$

$45-20 =(E_0 - 20) e^{-5k}$

Dividing,

$\dfrac{50}{25}=e^{4k}$

$k = \dfrac 1 4 \ln 2$

Now,

$E(t) - 20 = (E_0 - 20) e^{- k t}$

$E_0 = 20 + (E(t) - 20)e^{k t}$

$E_0 = 20 + (E(1) - 20)e^{k}$

$E_0 = 20 + (70 - 20)e^{\ln 2/4}$

$E_0 = 20 + 50 \cdot 2^{1/4} \approx 79.46 \textrm{ degrees F}$

3 0

3 years ago

Consider the probability that exactly 90 out of 148 students will pass their college placement exams. Assume the probability tha

Pepsi [2]

Answer:

0.0491 = 4.91% probability that exactly 90 out of 148 students will pass their college placement exams.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Assume the probability that a given student will pass their college placement exam is 64%.

This means that $p = 0.64$

Sample of 148 students:

This means that $n = 148$

Mean and standard deviation:

$\mu = E(X) = np = 148(0.64) = 94.72$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{148*0.64*0.36} = 5.84$

Consider the probability that exactly 90 out of 148 students will pass their college placement exams.

Due to continuity correction, 90 corresponds to values between 90 - 0.5 = 89.5 and 90 + 0.5 = 90.5, which means that this probability is the p-value of Z when X = 90.5 subtracted by the p-value of Z when X = 89.5.

X = 90.5

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

$Z = \frac{90.5 - 94.72}{5.84}$