5(x+2y)−3=3x+5 4(x−3y)−50=−33y X=? Y=?

5.44nL to mL convert

aniked [119]

$5.44nl \times \frac{1ml}{1000000nl} = 0.00000544ml$

6 0

3 years ago

What’s the answer? Why? Show the steps on how you go it

sleet_krkn [62]

This is not answer but Photomath is really helpful and it shows you step by step on how to solve it as well

7 0

3 years ago

50 pts please help me due today links or incorrect answers will get reported

Bumek [7]

Using the normal distribution, the area underneath the shaded region between the two z-scores is given by:

C. 0.6766.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

Hence, for this problem, the area is the p-value of z = 0.75 subtracted by the p-value of z = -1.3.

Looking at the z-table, the p-values are given as follows:

z = 0.75: 0.7734.

z = -1.3: 0.0968.

Then:

0.7734 - 0.0968 = 0.6766.

Which means that option C is correct.

More can be learned about the normal distribution at brainly.com/question/15181104

#SPJ1

3 0

1 year ago

FIND THE MEDIAN OF THE DATA SET3,35,23,37,45,49,27,48 A.35 B.34 C.30.2 D.38

mojhsa [17]

Answer:

A. 35

Step-by-step explanation:

The median of a data set is the middle value when the data values are placed in order of size.

Given data set:

{3, 35, 23, 37, 45, 5, 49, 27, 48}

Place the data in order of size:

{3, 5, 23, 27, 35, 37, 45, 48, 49}

To find the median, divide the total number of data values (n) by 2.

If n/2 is a whole number, the median is halfway between the values in this position and the position above.
If n/2 is not a whole number, round it up to find the position of the median.

As there are 9 data values, the median value is:

$\implies \dfrac{9}{2}=4.5 \implies \sf 5th\;value$

Therefore, the median of the given data set is 35.

4 0

1 year ago

Read 2 more answers

Two opposing opinions were shown to a random sample of 1,500 buyers of a particular political news app in the United States. The

kifflom [539]

Answer:

A) 98% Confidence interval for the proportion of all US buyers of this particular app who would have chosen Opinion B

= (0.51, 0.57)

This means that the true proportion of all thay would chose opinion B can take on values between the range of (0.51, 0.57)

B) For the confidence interval obtained to be valid, the conditions stated must be satisfied and for the sampling distribution to be approximately normal, the number of buyers that chose Opinion B and the number of buyers that did not choose Opinion B must both be greater than 10.

Step-by-step explanation:

Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample proportion) ± (Margin of error)

Sample proportion of all US buyers of this particular app who would have chosen Opinion B = 0.54

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value at 98% confidence interval for sample size of 1500 is obtained from the z-tables.

Critical value = 2.33

Standard error = σₓ = √[p(1-p)/n]

p = sample proportion = 0.54

n = sample size = 1500

σₓ = √(0.54×0.46/1500) = 0.0128685664 = 0.01287

98% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.54 ± (2.33 × 0.01287)

CI = 0.54 ± 0.02998)

98% CI = (0.51, 0.57)

98% Confidence interval = (0.51, 0.57)

B) The number of buyers that chose Opinion B and the number of buyers that did not choose Opinion B are both greater than 10. Why must this inference condition be met?

For this confidence interval to be obtained using sample data, a couple of conditions are necessary to be satisfied. They include;

- The sample data must have been obtained using a random sampling technique.

- The sampling distribution must be normal or approximately normal.

- The variables of the sample data must be independent of each other.

On the second point, the condition for a binomial distribution to approximate a normal distribution is that

np ≥ 10

and n(1-p) ≥ 10

The quantity np is the actual sample mean which is the actual number of buyers that chose Opinion B while n(1-p) is the number of buyers that did not chose Opinion B.

For the confidence interval obtained to be valid, the conditions stated must be satisfied and for the sampling distribution to be approximately normal, the number of buyers that chose Opinion B and the number of buyers that did not choose Opinion B must both be greater than 10.

Hope this Helps!!!

8 0

3 years ago