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2 years ago

7

The temperature of a cup of coffee decreased 26% in the first 5 minutes after it was made. The coffee starts at 190°F. What temp

erature is it after 5 minutes?

1 answer:

OLEGan [10]2 years ago

3 0

Answer:

it is 140.5 degrees after 5 min

You might be interested in

If p and q are both true, then which of the following statements has the same truth-value as ~p → q?

olganol [36]

If p and q are both true, then

$\neg p \implies q$

is an implication of the form

$F \implies T$

which is true, because every implication starting with false is true, i.e.

$F \implies T = T,\quad F \implies F = T$

So, we're looking for an expression evaluating to true. Let's see what we have:

A) is an AND proposition. Logical AND is true only if both parts are true. So, you have

$\neg P \land \neg Q = F \land F = F$

So it's not the right option.

B) is an OR proposition. Logical OR is true whenever one of the two parts is true. So, you have

$\neg P \lor\neg Q = F \lor F = F$

So it's not the right option.

C) is again an AND proposition. You have

$P \land \neg Q = T \land F = F$

So this is not the right option.

D) Finally, the last one is again an implication, and again it starts with false:

$\neg Q \implies P = F \implies T = T$

So this is true, and thus is the correct option.

8 0

2 years ago

Oliver has 3 times as many rocks in his collection as Katie. Then Oliver collected 75 more rocks and Katie collected 105 more ro

Evgesh-ka [11]

Answer:

The answer to your question is Katie had 15 rocks and Oliver 45 rocks.

Step-by-step explanation:

Conditions

Katie has x amount of rocks

Oliver has 3x amount of rocks

The final amount of rocks

Oliver = 3x + 75

Katie = x + 105

Now, they have the same amount of rocks, then we can equal both equations

3x + 75 = x + 105

Solve for x

3x - x = 105 - 75

2x = 30

x = 30/2

x = 15

Conclusions

At first Katie had 15 rocks and Oliver had 3(15) = 45 rocks

5 0

2 years ago

Read 2 more answers

The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5.

Natasha_Volkova [10]

Answer:

Yes, it would be unusual.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

If $Z \leq -2$ or $Z \geq 2$ , the outcome X is considered unusual.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 22, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1$

Would it be unusual for this sample mean to be less than 19 days?

We have to find Z when X = 19. So

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

By the Central Limit Theorem

$Z = \frac{19 - 22}{1}$

$Z = -3$

$Z = -3 \leq -2$ , so yes, the sample mean being less than 19 days would be considered an unusual outcome.

7 0

2 years ago

kyle unpacks a box that holds 9 layers of cans. There are 5 rows of 4 cans in each later. which way can Kyle organize all the ca

Ainat [17]

It might be 180 because 5×4 is 20 and 20×9 is 180

4 0

2 years ago

Read 2 more answers

A quantity x has cumulative distribution function p(x) = x − x2/4 for 0 ≤ x ≤ 2 and p(x) = 0 for x < 0 and p(x) = 1 for x &gt

34kurt

$F_X(x)=\begin{cases}0&\text{for }x2\end{cases}$

Recall that the PDF is given by the derivative of the CDF:

$f_X(x)=\dfrac{\mathrm dF_X(x)}{\mathrm dx}=\begin{cases}1-\dfrac x2\\\\0&\text{otherwise}\end{cases}$

The mean is given by

$\mathbb E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\int_0^2\left(x-\dfrac{x^2}2\right)\,\mathrm dx=\frac23$

The median is the number $M$ such that $F_X(x)=\mathbb P(X\le M)=\dfrac12$ . We have

$F_X(M)=M-\dfrac{M^2}4=\dfrac12\implies M=2\pm\sqrt2$

but both roots can't be medians. As a matter of fact, the median must satisfy $0\le M\le2$ , so we take the solution with the negative root. So $M=2-\sqrt2$ is the median.

3 0

3 years ago