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3 years ago

2. Your family eats at a Foo Fight Restaurant. The bill is $168. The family leaves a tip and spends a total of $201.60.

Mathematics

2 answers:

inysia [295]3 years ago

5 0

$33.6 in tips 201.60-168
20% tip 168x.2

Maksim231197 [3]3 years ago

3 0

Answer- A- 33.6
B- 20%
For A subtract subtract 168 from 201.60
For question B do 33.6÷168 and multiply it by 100

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Which of these numbers is between 5/6 and 1?<br> (A) 4/5 <br> (B) 4/7 <br> (C)6/7

djverab [1.8K]

Two basic ways in which to do this problem:

1. Find and apply the LCD.

2. Convert all of the given numbers to their decimal form, to 2 or 3 places only.

Try #2 first:

5/6 = 0.83

4/5 = 0.80 is not between 5/6 and 1. Reject it.
4/7 = 0.57 Reject
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3 years ago

Where were the two cars in relation to each other when they began traveling?

Elza [17]

Answer:

The answer is A.

Step-by-step explanation:

As can be seen in the graph, car B starts at the "20" mark on the vertical axis, signifying that it starts 20 miles east of car A.

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3 years ago

Rewrite 2 2/5 as an improper fraction. helpp

vesna_86 [32]

Answer: 12\5 :)

Step-by-step explanation: :)

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3 years ago

Read 2 more answers

A research study uses 800 men under the age of 55. Suppose that 30% carry a marker on the male chromosome that indicates an incr

ss7ja [257]

Answer:

a) There is a 12.11% probability that exactly 1 man has the marker.

b) There is a 85.07% probability that more than 1 has the marker.

Step-by-step explanation:

There are only two possible outcomes: Either the men has the chromosome, or he hasn't. So we use the binomial probability distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

30% carry a marker on the male chromosome that indicates an increased risk for high blood pressure, so $\pi = 0.30$

(a) If 10 men are selected randomly and tested for the marker, what is the probability that exactly 1 man has the marker?

10 men, so $n = 10$

We want to find P(X = 1). So:

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

$P(X = 1) = C_{10,1}.(0.30)^{1}.(0.7)^{9} = 0.1211$

There is a 12.11% probability that exactly 1 man has the marker.

(b) If 10 men are selected randomly and tested for the marker, what is the probability that more than 1 has the marker?

That is $P(X > 1)$

We have that:

$P(X \leq 1) + P(X > 1) = 1$

$P(X > 1) = 1 - P(X \leq 1)$

We also have that:

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

$P(X = 0) = C_{10,0}.(0.30)^{0}.(0.7)^{10} = 0.0282$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0282 + 0.1211 = 0.1493$

Finally

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1493 = 0.8507$

There is a 85.07% probability that more than 1 has the marker.

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3 years ago

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riadik2000 [5.3K]

Answer:

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