Please help me on this question and please explain it thanks

Solve the inequality for w. 20 ≥ - 4w Simplify your answer as much as possible.

Oksanka [162]

Answer:

w $\geq$ -5

Step-by-step explanation:

-4w $\leq$ 20

4w $\geq$ -20

w $\geq$ -5

8 0

3 years ago

1.Solve for w.

iren [92.7K]

Hi there,

1)
6 + (w - 10) = 5
6 + w - 10 = 5
- 4 + w = 5
w = 5 + 4

Hence, W = 9

2)
2/3 + y - 1/9 = 7/9
5/9 + y = 7/9
y= 7/9 - 5/9

Hence, Y = 2/9

4)
4x - 3/2x - 15/4 = 3/8
32x - 12x - 30 = 3
20x - 30 = 3
20x = 3 + 30
20x = 33

Hence, the answer is 33/20

5)
0.7(3s + 4) - 1.1s = 7.9
2.1s + 2.8 - 1.1 = 7.9
s + 2.8 = 7.9
s = 7.9 - 2.8

Hence, the answer is 5.1

Hope this all helps :)

8 0

3 years ago

Read 2 more answers

According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe

Ann [662]

Using the normal distribution and the central limit theorem, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$ , as long as $np \geq 10$ and $n(1 - p) \geq 10$ .

In this problem:

22% of couples meet online, hence p = 0.22.

A sample of 150 couples is taken, hence n = 150.

Item a:

The mean and the standard error are given by:

$\mu = p = 0.22$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{150}} = 0.0338$

The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

Item b:

The probability is one subtracted by the p-value of Z when X = 0.25, hence:

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

By the Central Limit Theorem:

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

$Z = \frac{0.25 - 0.22}{0.0338}$

Z = 0.89

Z = 0.89 has a p-value of 0.8133.

1 - 0.8133 = 0.1867.

There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

Item c:

The probability is the p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15, hence:

X = 0.2:

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

$Z = \frac{0.2 - 0.22}{0.0338}$

Z = -0.59

Z = -0.59 has a p-value of 0.2776.

X = 0.15:

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

$Z = \frac{0.15 - 0.22}{0.0338}$

Z = -2.07

Z = -2.07 has a p-value of 0.0192.

0.2776 - 0.0192 = 0.2584.

There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

To learn more about the normal distribution and the central limit theorem, you can check brainly.com/question/24663213

4 0

2 years ago

What is the equation of the function shown in the graph?

Zanzabum

Answer:

Step-by-step explanation:

I do not now what you mean?

5 0

2 years ago

The table shows how the number of sit-ups marla does each day has changed over time. at this rate how many sit-ups will she do o

DENIUS [597]

The number of situps form an AP with first term (a) = 17 and common difference (d) = 4

Tn = a + (n - 1)d
T12 = 17 + (12 - 1) x 4 = 17 + 11(4) = 17 + 44 = 61

Therefore, she will do 61 situps on day 12.

5 0

2 years ago

Read 2 more answers