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3 years ago

10

Mark and his three friends ate out at Applebee’s. Their bill totaled $104.70. If they left the server a 20% tip, how much would

each person pay splitting the bill evenly?

2 answers:

masya89 [10]3 years ago

6 0

It’s will be 31.41 dollars

natima [27]3 years ago

3 0

Answer:

$31.41!

Step-by-step explanation:

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What is -3x^2-4x-4=0

ss7ja [257]

Answer:

<h2>no real solution</h2>

Step-by-step explanation:

$-3x^2-4x-4=0\qquad\text{change the signs}\\\\3x^2+4x+4=0\\\\\text{use the quadratic formula:}\\\\\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac0,\ \text{then an equation has two solutions:}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\text{We have}\ a=3,\ b=4,\ c=4.\\\\\text{Substitute:}\\\\b^2-4ac=4^2-4(3)(4)=16-48=-32$

8 0

3 years ago

Given the box plot, will the mean or the median provide a better description of the center? box plot with min at 11, Q1 at 22. 5

Slav-nsk [51]

The median, because the data distribution is skewed to the right.

Given that

The box plot with the following parameters:

Minimum value at 11,

First Quartile, Q1 at 22.5,

Median at 34.5,

Third Quartile, Q3 at 36,

Maximum value at 37.5.

According to the question

When a distribution has a longer tail to one of its sides, then it is said to be skewed.

Since the tail is longer to the left, it is said to be left-skewed.

The data distribution is skewed to the left because the median (34.5) is closer to the third quartile (36) than to the first quartile (22.5).

For skewed data, we always prefer the median because it is less affected by the outliers and if the mean is chosen, the value of the mean is biased towards the side that has larger values while the median does not get affected by it. So, we choose the median.

For skewed distributions, the median is the best measure of center.

Hence, the median, because the data distribution is skewed to the right.

To know more about Mean click the link given below.

brainly.com/question/2508120

3 0

2 years ago

70 points I need help ASAP on everything but part A (WILL GIVE BRAINLIEST) Roller Coaster Crew Ray and Kelsey have summer intern

Airida [17]

7 is the right answer to the question

6 0

3 years ago

Read 2 more answers

You invest $500 in an account that has a annual interest rate of 5%, compounded quarterly for four years. What is the equivalent

posledela

Answer:

1.25% and 16 times

Step-by-step explanation:

Since the interest is compounded quarterly it will be compounded 4 times a year. So 4 x 4 is 16, so it will be compounded 16 times.

Then you have to divide the 5% by four to get how much will be compounded each quarter. So, (0.05 / 4) = 0.0125, which is 1.25%.

3 0

3 years ago

Assume that there is a 4% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive wit

kap26 [50]

Answer:

a) 99.84% probability that during a year, you can avoid catastrophe with at least one working drive

b) 99.999744% probability that during a year, you can avoid catastrophe with at least one working drive

Step-by-step explanation:

For each disk drive, there are only two possible outcomes. Either it works, or it does not. The disks are independent. So we use the binomial probability distribution to solve this question.

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.

4% rate of disk drive failure in a year.

This means that 96% work correctly, $p = 0.96$

a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk drive, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

This is $P(X \ geq 1)$ when $n = 2$

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.96)^{0}.(0.04)^{2} = 0.0016$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0016 = 0.9984$

99.84% probability that during a year, you can avoid catastrophe with at least one working drive

b. If copies of all your computer data are stored on four independent hard disk drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

This is $P(X \ geq 1)$ when $n = 4$

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{4,0}.(0.96)^{0}.(0.04)^{4} = 0.00000256$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.00000256 = 0.99999744$

99.999744% probability that during a year, you can avoid catastrophe with at least one working drive

7 0

3 years ago