All categories

3 years ago

-2 (h - 5) < -12 Please some help me I don’t understand :(

Mathematics

1 answer:

Marysya12 [62]3 years ago

6 0

Answer:

$h>11$

Step-by-step explanation:

We have the inequality:

$-2(h-5)$

Solve. Divide both sides by -2. Since we are dividing by a negative, we must flip the inequality sign. Thus:

$h-5>6$

Add 5 to both sides:

$h>11$

Therefore, our solution is all numbers greater than 11 .

This means that if we substitute any number greater than 11 into our original inequality, the inequality will be true.

You might be interested in

Find the sum.<br>(w - 2.4) + (1 - 0.5w)

vodka [1.7K]

The sum of the given expression { (w - 2.4) + (1 - 0.5w) } is 0.5w - 1.4.

<h3>What is the sum of the given expression?</h3>

Given the expression in the question;

(w - 2.4) + (1 - 0.5w)

Remove the parenthesis

w - 2.4 + 1 - 0.5w

Collect like terms and simplify

w - 2.4 + 1 - 0.5w

w - 0.5w - 2.4 + 1

1w - 0.5w - 2.4 + 1

0.5w - 1.4

Therefore, the sum of the given expression { (w - 2.4) + (1 - 0.5w) } is 0.5w - 1.4.

Learn more algebra Problems here; brainly.com/question/723406

#SPJ1

8 0

2 years ago

There are 20 boys and 18 girls in a class find the percentage of {1} boys {2} girls, in the class

Viktor [21]

Answer:

1. 52.63% 2. 47.36%

Step-by-step explanation:

20 + 18 = 38 total people

20 divided by 38 = 0.5263 to 52.63%

18 divided by 38 = 0.4736 to 47.36%

Hope this helps!

5 0

2 years ago

A number is multiplied by 5. If 12 is added to the product you get 37.

Korolek [52]

5x+12=37

5x=25

x=5

I hope this helps

5 0

4 years ago

The marketing director of a large department store wants to estimate the average number of customers who enter the store every f

Dahasolnce [82]

Answer:

$36.5674\leq x'\leq61.4326$

Step-by-step explanation:

If we assume that the number of arrivals is normally distributed and we don't know the population standard deviation, we can calculated a 95% confidence interval to estimate the mean value as:

$x-t_{\alpha /2}\frac{s}{\sqrt{n} } \leq x'\leq x-t_{\alpha /2}\frac{s}{\sqrt{n} }$

where x' is the population mean value, x is the sample mean value, s is the sample standard deviation, n is the size of the sample, $\alpha$ is equal to 0.05 (it is calculated as: 1 - 0.95) and $t_{\alpha /2}$ is the t value with n-1 degrees of freedom that let a probability of $\alpha/2$ on the right tail.