Solve this equation. 23x−15x=x−1 A. x=11315 B. x=178 C. x=1715 D. x=78

If you bike for 60 kilometers at 30 kilometers per hour and return at 20 kilometers per hour, what is the average speed for the

matrenka [14]

V1=30km/hr
v2=20km/hr
v=2v1v2/(v1+v2)
=2*30*20/(30+20)
=24km/hr

8 0

3 years ago

Consider a normal distribution curve where the middle 85 % of the area under the curve lies above the interval ( 8 , 14 ). Use t

NeTakaya

Answer:

$\mu = 11$

$\sigma = 2.08$

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.

Middle 85%.

Values of X when Z has a pvalue of 0.5 - 0.85/2 = 0.075 to 0.5 + 0.85/2 = 0.925

Above the interval (8,14)

This means that when Z has a pvalue of 0.075, X = 8. So when $Z = -1.44, X = 8$ . So

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

$-1.44 = \frac{8 - \mu}{\sigma}$

$8 - \mu = -1.44\sigma$

$\mu = 8 + 1.44\sigma$

Also, when X = 14, Z has a pvalue of 0.925, so when $X = 8, Z = 1.44$

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

$1.44 = \frac{14 - \mu}{\sigma}$

$14 - \mu = 1.44\sigma$

$1.44\sigma = 14 - \mu$

Replacing in the first equation

$\mu = 8 + 1.44\sigma$

$\mu = 8 + 14 - \mu$

$2\mu = 22$

$\mu = \frac{22}{2}$

$\mu = 11$

Standard deviation:

$1.44\sigma = 14 - \mu$

$1.44\sigma = 14 - 11$

$\sigma = \frac{3}{1.44}$

$\sigma = 2.08$

7 0

3 years ago

Let x denote the courtship time for a randomly selected female–male pair of mating scorpion flies (time from the beginning of in

gavmur [86]

Complete Question:

a) Is it plausible that X is normally distributed?

b) For a random sample of 50 such pairs, what is the (approximate) probability that the sample mean courtship time is between 100 min and 125 min?

Answer:

a) It is plausible that X is normally distributed

b) probability that the sample mean courtship time is between 100 min and 125 min is 0.5269

Step-by-step explanation:

a)X denotes the courtship time for the scorpion flies which indicates that is a real - valued random variable, and since normal distribution is a continuous probability distribution for a real valued random variable, it is plausible that X is normally distributed.

b) Probability that the sample mean courtship time is between 100 min and 125 min

$\mu = 120\\n = 50$

$P(x_{1} < \bar{X} < x_{2} ) = P(z_{2} < \frac{x_{2}- \mu }{SD} ) - P(z_{1} < \frac{x_{2}- \mu }{SD})$

$SD = \sqrt{\frac{\sigma^{2} }{n} } \\SD = \sqrt{\frac{110^{2} }{50} } \\SD = 15.56$

$P(100 < \bar{X}$

From the probability distribution table:

$P(z_{2} < 0.32 ) = 0.6255\\ P(z_{1} < -1.29) = 0.0986$

$P(100 < \bar{X}$

6 0

3 years ago

A bag has 2 blue marbles 3 red marbles and 5 white marbles. Which event has a probability greater than 1/5 select three options

Evgesh-ka [11]

Answer:

all

Step-by-step explanation:

all of them has the probability

4 0

3 years ago

A rectangular stirp of asphalt paving is 5 meters longer than it is wide. its area is 300 square meters. find the length and wid

kow [346]

Let's call L the width of the rectangle and W its width. The area of the rectangle is the product between the length and the width, and we are also told that the area is 300 square meters, so we can write
$A=L\cdot W=300$
Moreover, we know that the length is 5 meters longer than the width:
$L=W+5$
We have a system of 2 equations in 2 unknown variables, L and W. If we substitute the second equation into the first one, we get
$(W+5)\cdot W=300$
$W^2+5W-300=0$
which has two solutions: W=-20 and W=15. We can discard the negative solution since it does not have physical meaning, and now we can substitute the value of W into the second equation to find L:
$L=W+5=15+5=20$
Therefore, the rectangle has width 15 meters and length 20 meters.

7 0

3 years ago

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