Jared owed his brother

Here’s the answer

VLD [36.1K]

Answer:

thx bb

Step-by-step explanation:

5 0

3 years ago

What is the value of the expression?

Elza [17]

-(5 + (-11.3)) + (-4.3)

-(5 - 11.3) + (-4.3)

Distribute -1 into the parenthesis:

-5 + 11.3 + (-4.3)

-5 + 11.3 - 4.3

Add:

6.3 - 4.3

Subtract:

2

8 0

4 years ago

A standard deck of has 52 cards, and contains 13 cards of each of the four suits: diamonds, spades, hearts, and clubs. What is t

Verizon [17]

Answer: $\frac{13}{51}$

Step-by-step explanation:

Given: Number of cards in a standard deck = 52

If we already chosen a diamond card, then the number of cards left in the deck =52-1=51

The number of spades remains same = 13

Now, the probability that you choose a spade, given that you have already chosen a diamond is given by :-

$P(Spade|diamond)=\frac{13}{51}$

Hence, the the probability that you choose a spade, given that you have already chosen a diamond $=\frac{13}{51}$

7 0

3 years ago

Time spent using e-mail per session is normally distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that

liq [111]

Answer:

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know 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.

Central limit theorem

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