Which function could be shown in the graph below

What would be Kelly’s weight be in newtons if her mass is 70 kilograms?

notsponge [240]

A. 70 × g
B. 70/g
C. 70/g2
D. 70 × g2
E. 70

answer= A

4 0

1 year ago

A really bad carton of eggs contains spoiled eggs. An unsuspecting chef picks eggs at random for his ""Mega-Omelet Surprise."" F

Dima020 [189]

Answer:

(a) The probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b) The probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c) The probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

Step-by-step explanation:

The complete question is:

A really bad carton of 18 eggs contains 8 spoiled eggs. An unsuspecting chef picks 5 eggs at random for his “Mega-Omelet Surprise.” Find the probability that the number of unspoiled eggs among the 5 selected is

(a) exactly 5

(b) 2 or fewer

(c) more than 1.

Let X = number of unspoiled eggs in the bad carton of eggs.

Of the 18 eggs in the bad carton of eggs, 8 were spoiled eggs.

The probability of selecting an unspoiled egg is:

$P(X)=p=\frac{10}{18}=0.556$

A randomly selected egg is unspoiled or not is independent of the others.

It is provided that a chef picks 5 eggs at random.

The random variable X follows a Binomial distribution with parameters n = 5 and p = 0.556.

The success is defined as the selection of an unspoiled egg.

The probability mass function of X is given by:

$P(X=x)={5\choose x}(0.556)^{x}(1-0.556)^{5-x};\ x=0,1,2,3...$

(a)

Compute the probability that of the 5 eggs selected exactly 5 are unspoiled as follows:

$P(X=5)={5\choose 5}(0.556)^{5}(1-0.556)^{5-5}\\=1\times 0.05313\times 1\\=0.0531$

Thus, the probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b)

Compute the probability that of the 5 eggs selected 2 or less are unspoiled as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

$=\sum\imits^{2}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=0.0173+0.1080+0.2706\\=0.3959$

Thus, the probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c)

Compute the probability that of the 5 eggs selected more than 1 are unspoiled as follows:

P (X > 1) = 1 - P (X ≤ 1)

= 1 - P (X = 0) - P (X = 1)

$=1-\sum\limits^{1}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=1-0.0173-0.1080\\=0.8747$

Thus, the probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

6 0

1 year ago

Fill in the gaps in the arithmetic sequence -3, _ ,_ , _, _, _ 12

lord [1]

Answer:

-0.5, 2, 4.5, 7, 9.5

Step-by-step explanation:

The given terms are 6 apart, so the common difference is 1/6 of their difference:

d = (12 -(-3))/6 = 15/6 = 5/2 = 2.5

Add 2.5 to each term to get the next one. Then the sequence is ...

-3, -0.5, 2.0, 4.5, 7.0, 9.5, 12

8 0

2 years ago

Difference of 86.42 - 2.1.

densk [106]

Answer:

The first one

Step-by-step explanation:

The decimals line up when you subtract them

3 0

2 years ago

Read 2 more answers

To plan the budget for next year a college must update its estimate of the proportion of next year's freshmen class that will ne

Naily [24]

Answer:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

$z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491$

$p_v =P(z>2.491)=0.0064$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of applicants who request financial aid is significantly higher than 0.35

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=67 represent the applicants who request financial aid

$\hat p=\frac{67}{150}=0.447$ estimated proportion of applicants who request financial aid

$p_o=0.35$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of applicants who request financial aid is higher than 0.35.:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>2.491)=0.0064$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of applicants who request financial aid is significantly higher than 0.35

5 0

2 years ago