Simplify the following expression using the distributive property:-3(4x-5)

jake spends 2 hours on each homework assignment and he had 3 assignments each day.orlando spend 2 hiurs on each homework assignm

Andrej [43]

Ok so u have to multiply for this one

Work *Jake*:

So he spends 2 hours on each assignment. He has 3 assignments so u have to do :
2*3=6
He spend 6 hours a day doing home work so now u have to multiply that by 30:
6*30=180

Work *Orlando*:

So he spends 2 hours on each homework. He has 2 sheets each do so u have to multiply 2 and 2:
2*2=4

He spends 4 hours a day doing homwork so u have to multiply it by 30:
4*30=120

Answer:

So Jake spends 130 hours
Orlando spends 120 hours

Add both together:
120+130=250

They spend 250 hours together on work in 30 days/

4 0

3 years ago

Read 2 more answers

A sports shop had 80

Fynjy0 [20]

Answer:

30% = -0.1

Step-by-step explanation:

6 0

3 years ago

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What term describes the set of all possible input values for a function

nexus9112 [7]

Answer:

The Domain.

Step-by-step explanation:

That is the Domain. For example the domain of the function √x is the set

{x : x ≥ 0}.

8 0

3 years ago

Question Help Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a

koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

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

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So $\mu = 5656, \sigma = 333.3$

(a) What proportion of light bulbs will last more than 6262 hours?

The pvalue of the z-score of $X = 6262$ is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.