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4 years ago

You work as a health inspector and must visit each of the 15 restaurants in town once each week.

Mathematics

1 answer:

PtichkaEL [24]4 years ago

8 0

Answer:

Step-by-step explanation:

The no of different ways in which these restaurants can be visited

= ¹⁵P₁₅

= 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

= 1.30 x 10¹²

b ) Each combination takes 1 week and 50 weeks in a year to work

No of years to take to try all the combination

= 1.30 x 10¹² / 50

= 2.615 x 10¹⁰ years .

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Nick starts with 20 milligrams of a radioactive substance. The amount of the substance decreases by 12 each week for a number of

Gre4nikov [31]

Answer:

what the answer

Step-by-step explanation:

6 0

3 years ago

Consider the function below. f(x)= x^3 + 2x^2 - x - 2 plot the x and y intercepts of the function

dimulka [17.4K]

In the Figure below is shown the graph of this function. We have the following function:

$f(x)=x^3+2x^2-x-2$

The $y-intercept$ occurs when $x=0$ , so:

$f(0)=(0)^3+2(0)^2-(0)-2=-2$

Therefore, the $y-intercept$ is the given by the point:

$\boxed{(0,-2)}$

From the figure we have three $x-intercepts$ :

$\boxed{P_{1}(-2,0)} \\ \boxed{P_{2}(-1,0)} \\ \boxed{P_{3}(1,0)}$

So, the $x-intercepts$ occur when $y=0$ . Thus, proving this:

$f(x)=x^3+2x^2-x-2 \\ \\ For \ P_{1}:\\ If \ x=-2, \ y=(-2)^3+2(-2)^2-(-2)-2=0 \\ \\ For \ P_{2}:\\ If \ x=-1, \ y=(-1)^3+2(-1)^2-(-1)-2=0 \\ \\ For \ P_{3}:\\ If \ x=1, \ y=(1)^3+2(1)^2-(1)-2=0$

7 0

3 years ago

Read 2 more answers

What is the volume (in cubic units) of a sphere with a radius of 18 units? Assume that π = 3.14 and round your answer to the nea

Nataly [62]

Answer:

The volume of sphere is $24416.64\ cm^3$ .

Step-by-step explanation:

We have,

Radius of a sphere is 18 units.

It is required to find the volume of sphere.

The formula of the volume of sphere in terms of radius is given by :

$V=\dfrac{4}{3}\pi r^3$

Plugging all values in above formula

$V=\dfrac{4}{3}\times 3.14\times (18)^3\\\\V=24416.64\ cm^3$

So, the volume of sphere is $24416.64\ cm^3$ .

7 0

3 years ago

Reagan scored 1140 on the SAT. The distribution of SAT scores in a reference population is normally distributed with mean 1000 a

krok68 [10]

Answer:

Jessie scored higher than Reagan.

Step-by-step explanation:

We are given that Reagan scored 1140 on the SAT. The distribution of SAT scores in a reference population is normally distributed with mean 1000 and standard deviation 100.

Jessie scored 30 on the ACT. The distribution of ACT scores in a reference population is normally distributed with mean 17 and standard deviation 5.

For finding who performed better on the standardized exams, we have to calculate the z-scores for both people.

1. Finding z-score for Reagan;

Let X = distribution of SAT scores

SO, X ~ Normal( $\mu=1000, \sigma^{2}=100^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 1000

$\sigma$ = standard deviation = 100

Now, Reagan scored 1140 on the SAT, that is;

z-score = $\frac{1140-1000}{100}$ = 1.4

2. Finding z-score for Jessie;

Let X = distribution of ACT scores

SO, X ~ Normal( $\mu=17, \sigma^{2}=5^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 17

$\sigma$ = standard deviation = 5

Now, Jessie scored 30 on the ACT, that is;

z-score = $\frac{30-17}{5}$ = 2.6

This means that Jessie scored higher than Reagan because Jessie's standardized score was 2.6, which is 2.6 standard deviations above the mean and Reagan's standardized score was 1.4, which is 1.4 standard deviations above the mean.

6 0

3 years ago

Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of

IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if n = 100.

(2) A Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable X following a Binomial distribution with parameters n and p.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.10=1$

Thus, a Normal approximation to binomial can be applied for population 1, if n = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

8 0

3 years ago