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3 years ago

A spinner numbered 1 through 10 is spun 3 times.

Mathematics

1 answer:

Allisa [31]3 years ago

6 0

The probability is 1/10 * 5/10 * 5/10, or 0.025.

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Parts being manufactured at a plant are supposed to weigh 65 grams. Suppose the distribution of weights has a Normal distributio

andrew-mc [135]

Answer:

0.64%.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 75 grams and a standard deviation of 22 grams.

This means that $\mu = 75, \sigma = 22$

Sample of 144:

This means that $n = 144, s = \frac{22}{\sqrt{144}} = 1.8333$

More than 80 or less than 70:

Both are the same distance from the mean, so we find one probability and multiply by 2.

The probability that it is less than 70 is the pvalue of Z when X = 70. So

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

By the Central Limit Theorem

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

$Z = \frac{70 - 75}{1.8333}$

$Z = -2.73$

$Z = -2.73$ has a pvalue of 0.0032

2*0.0032 = 0.0064.

0.0064*100% = 0.64%

The probability is 0.64%.

3 0

3 years ago

The pyramid shown has a square base that is 18 inches on each side. The slant height is 16 inches. What is the surface area of t

Alik [6]

The answer is 28sqinches Why The Surface Area of a pyramid is A=L×W×Have since the base is l×with combined it's now SA=BA×H
SA=18×16
SA=288sqinches.

4 0

3 years ago

URGENT: I need these problems answered and steps :/ Will give the first person BRAINLIEST and please show the steps and please p

hjlf

Answer:

1) -15/32

2) 2/15

3) -203.5

4) -3/2 = -1.5

5) 10/3 = 3 1/3

Step-by-step explanation:

For 1, you just multiply the numerators and denominators together (3 by 5, 4 by 8) and then because it is negative, add the negative sign.

For 2, you multiply again, but because there are 2 negative signs, it becomes positive.

For 3, you divide 81.4 by 0.4, you could also divide 814 by 4.

For 4, you keep change flip, so the equation would become -9/4 (-2 1/4=-9/4) and 3/2 would become 2/3. The equation would be (-9/4 * 2/3). Then, you would get -18/12 which is -3/2.

For 5, you just multiply 4/3 by 5/2 which is 20/6 and because 2 negatives, it becomes positive.

3 0

3 years ago

Read 2 more answers

What is 175.42 rounded to the nearest tenth?

weqwewe [10]

175.4 because 4 is the tenth place and 2 would round down

4 0

2 years ago

Evaluate the iterated integral 2 0 2 x sin(y2) dy dx. SOLUTION If we try to evaluate the integral as it stands, we are faced wit

nignag [31]

Answer:

Step-by-step explanation:

Given that:

$\int^2_0 \int^2_x \ sin (y^2) \ dy dx \\ \\ \text{Using backward equation; we have:} \\ \\ \int^2_0\int^2_0 sin(y^2) \ dy \ dx = \int \int_o \ sin(y^2) \ dA \\ \\ where; \\ \\ D= \Big\{ (x,y) | }0 \le x \le 2, x \le y \le 2 \Big\}$

$\text{Sketching this region; the alternative description of D is:} \\ D= \Big\{ (x,y) | }0 \le y \le 2, 0 \le x \le y \Big\}$

$\text{Now, above equation gives room for double integral in reverse order;}$

$\int^2_0 \int^2_0 \ sin (y^2) dy dx = \int \int _o \ sin (y^2) \ dA \\ \\ = \int^2_o \int^y_o \ sin (y^2) \ dx \ dy \\ \\ = \int^2_o \Big [x sin (y^2) \Big] ^{x=y}_{x=o} \ dy \\ \\= \int^2_0 ( y -0) \ sin (y^2) \ dy \\ \\ = \int^2_0 y \ sin (y^2) \ dy \\ \\ y^2 = U \\ \\ 2y \ dy = du \\ \\ = \dfrac{1}{2} \int ^2 _ 0 \ sin (U) \ du \\ \\ = - \dfrac{1}{2} \Big [cos \ U \Big]^2_o \\ \\ = - \dfrac{1}{2} \Big [cos \ (y^2) \Big]^2_o \\ \\ = - \dfrac{1}{2} cos (4) + \dfrac{1}{2} cos (0) \\ \\$

$= - \dfrac{1}{2} cos (4) + \dfrac{1}{2} (1) \\ \\ = \dfrac{1}{2}\Big [1- cos (4) \Big] \\ \\ = \mathbf{0.82682}$

5 0

2 years ago