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3 years ago

15

I need help fast thx

1 answer:

andre [41]3 years ago

7 0

T will travel 81 m/s because addin six to 21 gives you 27 so take that times three and you get 81 m/s hope this helps

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(2x^2+4x+3) - (4x^2 -2x-3)

ankoles [38]

Answer:

x^{2}+6x+6 is your answer

7 0

3 years ago

Read 2 more answers

7 . Sampling distribution of the difference between two means A statistician is interested in the effectiveness of a weight-loss

UkoKoshka [18]

Answer:

0.229

Step-by-step explanation:

Given that the difference between the two sample means follows anormal distribution with a mean of11.00 and standard deviation equal to1.4387

A statistician is interested in the effectiveness of a weight-loss supplement. She randomly selects two independent samples. Individuals in the first sample of size n1 = 24 take the weight-loss supplement. Individuals in the second sample of size n2 = 21 take a placebo. Individuals in both samples follow identical exercise and diet programs. At the end of the study, the statistician measures the weight loss (in percent) of each participant.

We find that mean difference actual = 13-2 = 11

Probability that difference >12 =P(Z> $\frac{12-11}{1.3487} \\$ )

=P(Z>0.742)=.0.229

3 0

3 years ago

Suppose that a box contains 7 cameras and that 5 of them are defective. A sample of 2 cameras is selected at random without repl

irinina [24]

Answer:

E(x) = 1.43 (Approx)

Step-by-step explanation:

Given:

Total number of camera = 7

Defective camera = 5

Sample selected = 2

Computation:

when x = 0

P(x=0) = 2/7 × 1/6 = 2/42

P(x=1) = [2/7 × 5/6] + [5/7 × 2/6] = 20/42

P(x=2) = 5/7 × 4/6 = 20/42

So,

E(x) = [0×2/42] + [1×20/42] + [2×20/42]

E(x) = 1.43 (Approx)

8 0

3 years ago

How do I factor 5n^2 +19n+12

trapecia [35]

I hope this helps you

3 0

3 years ago

Let x denote the lifetime of a mcchine component with an exponential distribution. The mean time for the component failure is 25

aliina [53]

Answer:

0.1353 = 13.53% probability that the lifetime exceeds the mean time by more than 1 standard deviations

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

The mean time for the component failure is 2500 hours.

This means that $m = \frac{2500}, \mu = \frac{1}{2500} = 0.0004$

What is the probability that the lifetime exceeds the mean time by more than 1 standard deviations?

The standard deviation of the exponential distribution is the same as the mean, so this is P(X > 5000).

$P(X > x) = e^{-0.0004*5000} = 0.1353$

0.1353 = 13.53% probability that the lifetime exceeds the mean time by more than 1 standard deviations

4 0

2 years ago