How do i solve 14x-2=8x+6 (11+x)

Which measure of central tendency does 4 represent for the data shown?

kondor19780726 [428]

Answer:

4 is the median

Step-by-step explanation:

The mean is the average, the median is the middle value when listed from smallest to largest, and the mode is the value that appears most often

4 is the middle value since they are listed from smallest to largest

4 is the median

7 0

3 years ago

Read 2 more answers

What is the squared root of 1,000,0000ypt612y

bearhunter [10]

Its 1000 hope it helps

5 0

3 years ago

A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as

Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c) The probability that at most five must be selected to find four that are not seconds is 0.9453.

Step-by-step explanation:

Let X = number of seconds in the batch.

The probability of the random variable X is, p = 0.31.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...$

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

$P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888$

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

$=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776$

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

$={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453$

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

8 0

3 years ago

The number of hours a lightbulb burns before failing varies from bulb to bulb. The population distribution of burnout times is s

liubo4ka [24]

ANSWER:

The average burnout time of a large number of bulbs has a sampling distribution that is close to Normal.

STEP-BY-STEP EXPLANATION:

The cental limit theorem states, that id the sample size is large (30 or more), then the sampling distribution of the sample means is approximately normal with mean ц and standar deviation б/ $\sqrt{n}$

Thus the correct answer is the average burnout time of a large number of bulbs has a sampling distribution that is close to Normal.

6 0

3 years ago

2. Sandi can run for one hour at a constant speed of 348 cm per second. Marci can run for one

Zanzabum

Answer:

Marci is the faster runner.

Step-by-step explanation:

Sandi can run 348 cm in a second. Let's find out how many cm she can run in a minute: in a minute there are 60 seconds. So lets multiply 348cm times 60, this gives you 20,880. That is how much she travels in a minute, now let's find out how much she runs in an hour: if she runs 20,880 cm in an minute and there are 60 minutes in an hour, then all you have to do is multiply 20,880 times 60, this gives you 1,252,800. That is how many cm she runs in an hour. Now let's change from cm to miles. There are 160,934 cm in a mile. If we divide 1,252,800 by 160,934 this will give you

7 0

1 year ago