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3 years ago

7

PLEASE PLEASE PLEASE HELP! WILL GET ZERO!!

1 answer:

ASHA 777 [7]3 years ago

5 0

Median, upper quartile, and maximum are all incorrect

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Will give Brainliest if correct!

iren [92.7K]

Answer:

Add 2 to each next number

2 plus 2 is 4

4 plus 2 is 6

6 plus 2 is 8

4 0

3 years ago

Read 2 more answers

PLEASE HELP!!!!!!!!!!

Yuki888 [10]

Answer:

240

Step-by-step explanation:

4 kinds of crust along with 5 different meats, 4 different cheeses, and 3 vegetables for toppings

Combos = 4*5*4*3

=240

5 0

3 years ago

Plz help me and plz show work

Lana71 [14]

There is no y intercept since there are no points on y axis so?

5 0

3 years ago

Read 2 more answers

A game starts by flipping a coin. If you get heads, then you get to roll a die and, if you roll a 4, you win the jackpot. If you

Mariana [72]

$\frac{5}{24}$

Step-by-step explanation:

There are two ways to win a jackpot.

Let $p_{1}$ be the probability for first method and $p_{2}$ be the probability for second method.

Probability to get a head= $p_{head}$ = $\frac{1}{2}$

Probability to get a $4$ in the die= $p_{4}$ $=\frac{1}{6}$

Probability to get a head and get a 4 in die=[ $p_{1}$ tex]p_{head}\times p_{4}[/tex]= $\frac{1}{2}\times \frac{1}{6} =\frac{1}{12}$

Probability to get a tail= $p_{tail}$ = $\frac{1}{2}$

Probability to get a heart= $p_{heart}$ = $\frac{13}{52} =\frac{1}{4}$

Probability to get a tail and get a heart in the deck= $p_{tail} \times p_{heart}$ = $\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$

Total probability to win a jackpot= $p_{1}+p_{2}=\frac{1}{12}+\frac{1}{8}=\frac{5}{24}$

So,probability to win a jackpot is $\frac{5}{24}$

3 0

4 years ago

The AAA system contains 3 independent components in parallel, each with probability of failure 0.4. Find the expected number of

Gre4nikov [31]

Answer:

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it fails, or it does not. The components are independent. We want to know how many outcomes until r failures. The expected value is given by

$E = \frac{r}{p}$

In which r is the number of failures we want and p is the probability of a failure.

In this problem, we have that:

r = 1 because we want the first failed unit.

$p = 0.4[\tex]So[tex]E = \frac{r}{p} = \frac{1}{0.4} = 2.5$

The expected number of systems inspected until the first failed unit is 2.5

8 0

3 years ago