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2 years ago

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Using a standard six-sided die, janice rolled either a 2 or a 5on the last 6 rolls of the die. what is the probability of janice

rolling a 2or a 5 on her 7th toss of the die?

1 answer:

Vedmedyk [2.9K]2 years ago

8 0

Answer:

1/3

Step-by-step explanation:

probability = favorable outcomes/total possible outcomes

There are two favorable outcomes, 2 and 5.

There are 6 total possible outcomes.

Make your fraction.

2/6

reduce

1/3

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if a store is offering 20% discount on all items. is there a proportional relationship between the original price of an item and

Novosadov [1.4K]

Answer:

A store is offering a 14% discount on all items. Write an equation relating the sale price S for an item to its list price L.

Step-by-step explanation:

8 0

3 years ago

Researchers claim that 8% of people have blue eyes. Suppose the researchers' claim is true. Mrs. Greene has a Geometry class wit

vovikov84 [41]

Answer:

Probability that exactly 5 of them have blue eyes is 0.1165.

Step-by-step explanation:

We are given that Researchers claim that 8% of people have blue eyes. Suppose the researchers' claim is true. Mrs. Greene has a Geometry class with 40 students.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 40 students

r = number of success = exactly 5

p = probability of success which in our question is % of people

having blue eyes, i.e; 8%

<em>LET X = Number of students having blue eyes</em>

So, it means X ~ $Binom(n= 40,p=0.08)$

Now, Probability that exactly 5 of them have blue eyes is given by = P(X = 5)

P(X = 5) = $\binom{40}{5}\times 0.08^{5} \times (1-0.08)^{40-5}$

= $658008 \times 0.08^{5} \times 0.92^{35}$

= 0.1165

Therefore, Probability that exactly 5 of them have blue eyes is 0.1165.

7 0

3 years ago

20% of US High School teens vape. A local High School has implemented campaigns to reduce vaping among students and believes tha

zaharov [31]

Answer:

10.93% probability of observing 51 or fewer vapers in a random sample of 300

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 300, p = 0.2$

So

$\mu = E(X) = np = 300*0.2 = 60$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.2*0.8} = 6.9282$

What is the approximate probability of observing 51 or fewer vapers in a random sample of 300?

Using continuity corrections, this is $P(X \leq 51 + 0.5) = P(X \leq 51.5)$ , which is the pvalue of Z when X = 51.5 So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{51.5 - 60}{6.9282}$

$Z = -1.23$

$Z = -1.23$ has a pvalue of 0.1093.

10.93% probability of observing 51 or fewer vapers in a random sample of 300

4 0

2 years ago

1.) If I am selling a dozen at $3, how much would each egg cost?

riadik2000 [5.3K]

Answer:

1.$3÷12=0.25

each egg would cost 25 cents

2.12×18=216

216×25=5,400

6 0

2 years ago

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify y

mariarad [96]

Answer:

the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;

$\frac{\pi }{2} [e^2 - 1 ]$ or 10.036

Step-by-step explanation:

Given the data in the question;

y = $y = e^{(x - 1 )$ , y = 0, x = 1, x = 2.

Now, using the integration capabilities of a graphing utility

y = $y = e_2}^{(x - 1 )_$ , y = 0

Volume = $\pi \int\limits^2_1 ( e^{x-1)^2} - (0)^2 dx$

Volume = $\pi \int\limits^2_1 ( e^{x-1)^2 dx$

Volume = $\pi \int\limits^2_1 e^{2x-2}dx$

Volume = $\frac{\pi }{e^2} \int\limits^2_1 e^{2x}dx$

Volume = $\frac{\pi }{e^2} [\frac{e^{2x}}{2}]^2_1$

Volume = $\frac{\pi }{2e^2} [e^4 - e^2 ]$

Volume = $\frac{\pi }{2} [e^2 - 1 ]$ or 10.036

Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;

$\frac{\pi }{2} [e^2 - 1 ]$ or 10.036

3 0

3 years ago