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3 years ago

A coin is tossed 5 times. let x count the number of heads tossed. determine x hhtht ( ).

2 answers:

8 0

Given that x counts the number of heads tossed

If the sequence of the results of the coin toss is HHTHT, it can be seen that there are 3 heads tossed.

Therefore, x(HHTHT) = 3

Maslowich3 years ago

4 0

Given that x counts the number of heads tossed

If the sequence of the results of the coin toss is HHTHT, it can be seen that there are 3 heads tossed.

Therefore, x(HHTHT) = 3 i that help u

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For the differential equation s′′+bs′+6s=0, find all the values of b that make the general solution overdamped, those that make

djyliett [7]

Answer:

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solution is attached

6 0

3 years ago

Please need help on this

Alex

Answer:

there is an 88% chance it will land on teal, because if you do 16/18, you get 0.8 repeating. then, you multiply by 100 and get 88%

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2 years ago

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

MrRissso [65]

Answer:

We conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

We conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

Step-by-step explanation:

We are given a random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen;

1.10, 5.09, 0.97, 1.59, 4.60, 0.32, 0.55, 1.45, 0.14, 4.47, 1.20, 3.50, 5.02, 4.67, 5.22, 2.69, 3.98, 3.17, 3.03, 2.21, 0.69, 4.47, 3.31, 1.17, 0.76, 1.17, 1.57, 2.62, 1.66, 2.05.

Let $\mu$ = true average percentage of organic matter

So, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is something other than 3%}

The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean percentage of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, the test statistics = $\frac{2.481-3}{\frac{1.616}{\sqrt{30} } }$ ~ $t_2_9$

= -1.76

The value of t-test statistics is -1.76.

(a) Now, at 10% level of significance the t table gives a critical value of -1.699 and 1.699 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

(b) Now, at 5% level of significance the t table gives a critical value of -2.045 and 2.045 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

8 0

3 years ago

The computer club spent $2,565 on mouse pads. The me never plan to sell the nous pads during the book fair. If they charge $9.50

Lapatulllka [165]

The answer is 270 because 2,565/9.5=270 hope this helps!

8 0

2 years ago

Number 9 please. I need help

Sloan [31]

(1/12)X = 6800

6800/ (1/12) = X

X = 81600 (hours)

6800 (6.8 * 10^3) divided by 1/12 (12^-1) equals 81600. It will take 81600 hours for the pool to completely empty at a rate of 0.833333333 or 12^-1 gallons per second.

8 0

3 years ago