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3 years ago

In tossing four fair dice, what is the probability of tossing, at most, one 3

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

7 0

Answer-

In tossing four fair dice, the probability of getting at most one 3 is 0.86.

Solution-

The probability of getting at most one 3 is, either getting zero 3 or only one 3.

$P(A) =The \ probability \ of \ getting \ zero \ 3 =(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})=\frac{625}{1296} =0.48$ ( ∵ xxxx )

$P(B) = The \ probability \ of \ getting \ only \ one \ 3 =(4)(\frac{1}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6}) = 0.38$ ( ∵ 3xxx, x3xx, xx3x, xxx3 )

P(Atmost one 3) = P(A) + P(B) = 0.48 + 0.38 = 0.86

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Lines x and y are parallel and are cut by the transversal b. If the measure of angle 1 is 57°, what is the measure of angle 6? E

lana [24]

Answer:

$\angle6=123$

Step-by-step explanation:

Given

Lines x and y

Transversal b

$\angle 1 = 57$

Required

Find $\angle 6$

From the attachment, $\angle1$ and $\angle7$ are vertically opposite.

This means that

$\angle 1 =\angle 7 = 57$

Similarly

$\angle 7$ and $\angle 6$ are supplementary angles

So:

$\angle7 + \angle6=180$

Make $\angle6$ the subject

$\angle6=180-\angle7$

$\angle6=180-57$

$\angle6=123$

The relationship between $\angle1$ and $\angle6$ is that they are supplementary angles because:

$\angle6+ \angle1=180$

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2 years ago

Allen�s hummingbird (Selasphorus sasin ) has been studied by zoologist Bill Alther A small group of 15 Allen� s hummingbirds has

Bogdan [553]

Answer:

1) 80% CI: [3.04; 3.26]gr

d= 0.11

2) n= 28 hummingbirds

Step-by-step explanation:

Hello!

The study variable of this experiment is:

X: the weight of a hummingbird. (gr)

And it has a normal distribution, symbolically: X~N(μ;σ²)

And (I hope I got it correctly) its population standard deviation is σ= 0.33

There was a sample of n= 15 hummingbirds taken, its sample mean X[bar]= 3.15 gr

1) You need to construct an 80% Confidence Interval for the population mean of the hummingbird's weight.

Since the study variable has a normal distribution, you can use either the standard normal distribution or the Student's t distribution. Both are useful to estimate the population mean. Since the population standard variance is known, the best choice is the Standard normal.

Z= X[bar] - μ ~ N(0;1)

σ/√n

The formula for the interval is:

X[bar] ± $Z_{1- \alpha /2}$ * (σ/√n)

$Z_{1- \alpha /2}= Z_{0.90} = 1.28$

3.15 ± 1.28 * (0.33/√15)

[3.04; 3.26]gr

The margin of error (d) of a confidence interval is hal its amplitude (a)

a= Upper bond - Lower bond

d= (Upper bond - Lower bond)/2

d= $\frac{(3.26-3.04)}{2}$ = 0.11

2) You need to calculate a sample size for a 80% Confidence interval for the average weight of the hummingbirds with a margin of error of d= 0.08

As I said before, the margin of error is half the amplitude of the interval, the formula you use to estimate the population mean has the following structure:

"point estamator" ± "margin of error"

Then the margin of error is:

d= $Z_{1- \alpha /2}$ * (σ/√n)

Now what you have to do is rewrite the formula based on the sample size

d= $Z_{1- \alpha /2}$ * (σ/√n)

$\frac{d}{Z_{1- \alpha /2}}$ = σ/√n

√n * $\frac{d}{Z_{1- \alpha /2}}$ = σ

√n = σ * $\frac{Z_1- \alpha /2}{d}$

n = (σ * $\frac{Z_1- \alpha /2}{d}$ )²

n= (0.33 * $\frac{1.28}{0.08}$ )²

n= 27.8784 ≅ 28 hummingbirds.

I hope it helps!

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3 years ago

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Answer:

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Then in a year Jim's dog would get

So Jim's dog will get 17- 13 =4 treatments more.

4 treatments that are made in 3 weeks each will be 4×3 =12 weeks more treatment

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