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3 years ago

Which expression is equivalent to 6e+3(e-1)

Mathematics

2 answers:

Dahasolnce [82]3 years ago

5 0

6e + 3(e - 1)

6e + 3e - 3

9e - 3

Katen [24]3 years ago

4 0

Answer: Equivalent expression would be 9e-3.

Step-by-step explanation:

Since we have given that

$6e+3(e-1)$

As we need to simplify the above expression:

First we open the brackets :

$3(e-1)=3e-3$

Now, add it to 6e.

So, it becomes,

$6e+3e-3\\\\=9e-3$

Hence, equivalent expression would be 9e-3.

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Need help with 6 questions will upvote

Black_prince [1.1K]

I dont need help sorry

8 0

3 years ago

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean u=207 and standard d

never [62]

Answer:

a) 36.32% probability that a randomly selected pregnancy lasts less than 201 days.

b) 3.22% probability that a random sample of 16 preganacies has a mean geatation period of 201 days or less.

c) 1.58% probability that a random sample of 37 preganacies has a mean geatation period of 201 days or less

Step-by-step explanation:

To solve this problem, we have to know the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit theorem

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

In this problem, we have that:

$\mu = 207, \sigma = 17$

(a) what is the probability that a randomly selected pregnancy lasts less than 201 days?

Pvalue of Z when X = 201.

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

$Z = \frac{201 - 207}{17}$

$Z = -0.35$

$Z = -0.35$ has a pvalue of 0.3632.

So 36.32% probability that a randomly selected pregnancy lasts less than 201 days.

(b) what is the probability that a random sample of 16 preganacies has a mean geatation period of 201 days or less?

Now $n = 16, s = \frac{17}{\sqrt{16}} = 3.25$

Again the pvalue of Z when X = 201

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

Due to the Central Limit Theorem

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

$Z = \frac{201 - 207}{3.25}$

$Z = -1.85$

$Z = -1.85$ has a pvalue of 0.0322.

So 3.22% probability that a random sample of 16 preganacies has a mean geatation period of 201 days or less.

(c) what is the probability that a random sample of 37 preganacies has a mean geatation period of 201 days or less?

Now $n = 16, s = \frac{17}{\sqrt{37}} = 2.79$

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

$Z = \frac{201 - 207}{2.79}$

$Z = -2.15$

$Z = -2.15$ has a pvalue of 0.0158.

1.58% probability that a random sample of 37 preganacies has a mean geatation period of 201 days or less

8 0

4 years ago

An astronaut weighs 170 pounds on Earth and 400 pounds on Jupiter. The astronaut captures an alien that weighs 120 pounds on Jup

Svet_ta [14]

B. 51 pounds.
<span>You multiply your weight on Earth (170) and the Alien's weight one Jupiter (120) and divide it by your weight on Jupiter (400).</span>

5 0

3 years ago

Using the following figure, determine the measure of

vampirchik [111]

3x = 108
devide by 3 to both side

x = 108 ÷ 3

x = 36°

option A is correct

7 0

3 years ago

If g(x) = /-83. find g(19)

irina [24]

The question doesn't make sense

g(19) means plug in 19 where you see an x, but your equation doesn't have any x in it

7 0

3 years ago

Which expression is equivalent to 6e+3(e-1)​

Which expression is equivalent to 6e+3(e-1)