Please help guys this is hard for my 3rd grader Brain

An unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6

SIZIF [17.4K]

Answer:

Probability of rolling at least 4 sixes is 0.01696.

Step-by-step explanation:

We are given that an unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6 times.

The above situation can be represented through binomial distribution;

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

where, n = number trials (samples) taken = 6 trials

r = number of success = at least 4

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

rolling a “six", i.e; p = 0.20

Let X = Number of sixes on a die

So, X ~ Binom(n = 6, p = 0.20)

Now, Probability of rolling at least 4 sixes is given by = P(X $\geq$ 4)

P(X $\geq$ 4) = P(X = 4) + P(X = 5) + P(X = 6)

= $\binom{6}{4} \times 0.20^{4} \times (1-0.20)^{6-4}+\binom{6}{5} \times 0.20^{5} \times (1-0.20)^{6-5}+\binom{6}{6} \times 0.20^{6} \times (1-0.20)^{6-6}$

= $15 \times 0.20^{4} \times 0.80^{2}+6 \times 0.20^{5} \times 0.80^{1}+1 \times 0.20^{6} \times 0.80^{0}$

= 0.0154 + 0.00154 + 0.000064

= 0.01696

Therefore, probability of rolling at least 4 sixes is 0.01696.

8 0

2 years ago

Timothy evaluated the expression using x = 3 and y = –4. xy-2 3x2y−4 1. (1 3 )x−1y2 2. (1 3 )3−1(−4)2 3. (1 3 )(1 31 )(−4)2 4. (

Sergio039 [100]

If this is for E2020 the answer is
D)No, his value of should be positive because an even exponent indicates a positive value.
thats why it was wrong

5 0

3 years ago

Read 2 more answers

PLEASE HELP ME! 100 POINTS! ;( question 10, 11, 12, 13, 14, 15. I have a f in this class and could use your help a lot! I will l

MatroZZZ [7]

There are a lot of problems here so I'll try to be brief with each one so I don't add a lot of clutter

Problem 10) This is arithmetic because we subtract 6 from each term to get the next. This is the same as adding on -6. The common difference is d = -6. The first term is a = 1

Problem 11) Neither. The distance from 3 to 3/2 = 1.5 is 1.5 units, but the distance from 1.5 to 1 is 0.5 units. This implies there is no common difference value. The sequence is not arithmetic because of this. It's also not geometric either. To go from 3 to 3/2, we multiply by 1/2. But then going from 3/2 to 1, we multiply by 2/3. There is no common ratio r value.

Problem 12) This sequence is geometric. We divide each term by 3 to get the next, or put another way, we multiply each term by 1/3 to get the next term. The common ratio is r = 1/3. The first term is a = 108.

Problem 13) Neither. This is because going from term to term, we do not add the same amount each time. Example: from -2 to 4 we add 6, but then from 4 to -6 we add -10. So the sequence is not arithmetic. It's also not geometric either because we dont multiply by the same term each time. Eg: from -2 to 4, we multiply by -2; but from 4 to -6 we multiply by -1.5

Problem 14) The next three terms are: 1, 1/4, 1/16. This is found by multiplying each term by the common ratio r = 1/4, or you can think of dividing each term by 4. To get this common ratio, pick any term you want and divide it by its previous term. Example: term2/term1 = 16/64 = 1/4 = common ratio.

Problem 15) The next three terms are: -432, 2592, -15552. You multiply each term by the common ratio -6. Like with the previous problem, we divide any term over its previous one to get the common ratio, so for example, term2/term1 = -12/2 = -6 = common ratio.

3 0

3 years ago

Am I correct or wrong on my answer

astraxan [27]

Answer:

wrong

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

A test of H0: p = 0.5 versus Ha: p > 0.5 has the test statistic z = 1.15. Part A: What conclusion can you draw at the 5% sign

Nadya [2.5K]

Answer:

Part A: The null hypothesis failed to be rejected.

Part B: The null hypothesis failed to be rejected.

Step-by-step explanation:

We have an hypothesis test with null and alternative hypothesis H0: p = 0.5 versus Ha: p > 0.5, which has the test statistic z=1.15.

Part A: If the significance level is 0.05, the conclusion depends on the P-value.

If the P-value is below 5%, the null hypothesis is rejected.

The P-value for this right-tailed tes and z=1.15 is:

$P-value=P(z>1.15)=0.125$

The P-value is bigger than the significance level, so the effect is not significant and the null hypothesis is failed to be rejected.

Part B: In this case, the significance level is 0.01 and, as the alternative hypothesis is defined with an unequal sign, the test is two-tailed.

This changes the way we calculate the P-value, as we need to compute the two tails.

The P-value is:

$P-value=2P(z>1.15)=0.25$

The P-value is bigger than the significance level, so the effect is not significant and the null hypothesis is failed to be rejected.

5 0

3 years ago