The cargo of the truck weighs at most 2,900 pounds. Use w to represent the weight (in pounds) of the cargo.

Solve the system by substitution. y = 4x + 22 y = -7x

Lesechka [4]

Answer:

x=-2

y=14

Step-by-step explanation:

-7x=4x+22

-11x=22

x=-2

y=-7(-2)

y=14

7 0

3 years ago

Read 2 more answers

Dave’s Automatic Door, referred to in Exercise 29, installs automatic garage door openers. Based on a sample, following are the

HACTEHA [7]

The question is not complete and the full question says;

Calculate the (a) range, (b) arithmetic mean, (c) mean deviation, and (d) interpret the values. Dave’s Automatic Door installs automatic garage door openers. The following list indicates the number of minutes needed to install a sample of 10 door openers: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42.

Answer:

A) Range = 30 minutes

B) Mean = 38

C) Mean Deviation = 7.2

D) This is well written in the explanation.

Step-by-step explanation:

A) In statistics, Range = Largest value - Smallest value. From the question, the highest time is 54 minutes while the smallest time is 24 minutes.

Thus; Range = 54 - 24 = 30 minutes

B) In statistics,

Mean = Σx/n

Where n is the number of times occurring and Σx is the sum of all the times occurring

Thus,

Σx = 28 + 32 + 24 + 46 + 44 + 40 + 54 + 38 + 32 + 42 = 380

n = 10

Thus, Mean(x') = 380/10 = 38

C) Mean deviation is given as;

M.D = [Σ(x-x')]/n

Thus, Σ(x-x') = (28-38) + (32-38) + (24-38) + (46-38) + (44-38) + (40-38) + (54-38) + (38-38) + (32-38) + (42-38) = 72

So, M.D = 72/10 = 7.2

D) The range of the times is 30 minutes.

The average time required to open one door is 38 minutes.

The number of minutes the time deviates on average from the mean of 38 minutes is 7.2 minutes

4 0

3 years ago

Multiply. Write your answer as a fraction in simplest form.

viva [34]

Answer:

no worries :)

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Question 2 of 10

RideAnS [48]

Answer: 264,615

Step-by-step explanation:apex

5 0

3 years ago

Radioactive Decay:

Vadim26 [7]

The question is incomplete, here is the complete question:

The half-life of a certain radioactive substance is 46 days. There are 12.6 g present initially.

When will there be less than 1 g remaining?

Answer: The time required for a radioactive substance to remain less than 1 gram is 168.27 days.

Step-by-step explanation:

All radioactive decay processes follow first order reaction.

To calculate the rate constant by given half life of the reaction, we use the equation:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half life period of the reaction = 46 days

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{46days}\\\\k=0.01506days^{-1}$

The formula used to calculate the time period for a first order reaction follows:

$t=\frac{2.303}{k}\log \frac{a}{(a-x)}$

where,

k = rate constant = $0.01506days^{-1}$

t = time period = ? days

a = initial concentration of the reactant = 12.6 g

a - x = concentration of reactant left after time 't' = 1 g

Putting values in above equation, we get:

$t=\frac{2.303}{0.01506days^{-1}}\log \frac{12.6g}{1g}\\\\t=168.27days$

Hence, the time required for a radioactive substance to remain less than 1 gram is 168.27 days.

7 0

3 years ago