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2 years ago

A 6 sided die is rolled 2 times. What is the probability of getting a 1 both times ?

Mathematics

1 answer:

STatiana [176]2 years ago

8 0

Answer:

Step-by-step explanation: If you roll the die twice, the probability of getting a even number both times is (1/2)(1/2) ... A probability experiment consists of rolling a 6 sided die

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A kid has 3 balls he gives his friends 2 how many do they have left

Olin [163]

Answer:

1 left

Step-by-step explanation:

3 - 2 =1 You subtract then u get your answer

Hope this help u

Pls mark brainlist

3 0

3 years ago

Read 2 more answers

Use a half-angle identity to find the exact value

Tatiana [17]

Given:

$\cos 15^{\circ}$

To find:

The exact value of cos 15°.

Solution:

$$\cos 15^{\circ}=\cos\frac{ 30^{\circ}}{2}$

Using half-angle identity:

$$\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos (x)}{2}}$

$$\cos \frac{30^{\circ}}{2}=\sqrt{\frac{1+\cos \left(30^{\circ}\right)}{2}}$

Using the trigonometric identity: $\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$$=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}$

Let us first solve the fraction in the numerator.

$$=\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}$

Using fraction rule: $\frac{\frac{a}{b} }{c}=\frac{a}{b \cdot c}$

$$=\sqrt{\frac {2+\sqrt{3}}{4}}$

Apply radical rule: $\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

$$=\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}$

Using $\sqrt{4} =2$ :

$$=\frac{\sqrt{2+\sqrt{3}}}{2}$

$$\cos 15^\circ=\frac{\sqrt{2+\sqrt{3}}}{2}$

5 0

4 years ago

I need help with 4,5,6

irina [24]

Answer:

4. 3. 10

5. 1102.31

6. 17.37

Step-by-step explanation:

5 0

3 years ago

Suppose a random variable, x, arises from a binomial experiment. If n = 25, and p = 0.85, find the variance. Round answer to 4 d

Tamiku [17]

Answer:

a) P(X=1) = 0.302526

b) P(X=5) = 0.010206

c) P(X=3) = 0.18522

d) P(X≤3) = 0.92953

e) P(X≥5) = 0.010935

f) P(X≤4) = 0.989065

If helpful, please mark as brainliest! =)

8 0

3 years ago

For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of sc

gogolik [260]

Answer:

a) The percentage of bone density scores that are significantly high is 2.28%

b) The percentage of bone density scores that are significantly low is 2.28%

c) The percentage of bone density scores that are not significant is 95.44%

Step-by-step explanation:

Problems of normally distributed samples are 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.

a. significantly high (or at least 2 standard deviations above the mean).

This is 1 subtracted by the pvalue of Z = 2.

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% of scores are signifcantly high

b. significantly low (or at least 2 standard deviations below the mean).

pvalue of Z = -2

Z = -2 has a pvalue of 0.0228

2.28% of scores are signicantly low.

c. not significant (or less than 2 standard deviations away from the mean).

pvalue of Z = 2 subtracted by the pvalue of Z = -2.

Z = 2 has a pvalue of 0.9772

Z = -2 has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

95.44% of the scores are not significant

7 0

4 years ago