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3 years ago

A truck is carrying 3 tons of cement and 150 pounds of bricks. How many pounds does the truck's load weigh in total?

Mathematics

2 answers:

Andreas93 [3]3 years ago

6 0

Answer:

the answer is 6150

Step-by-step explanation:

wariber [46]3 years ago

5 0

6150 not sure but if im wrong im so sorry

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How do you divide 100.5/1.5

Artist 52 [7]

It is: 1.5/100.5 = 1/67 4 goes into 9 two times with a remainder of 1. Bring down the 6 and 4 goes into 16 4 times.So 96 divided by 4 is 24

7 0

3 years ago

Read 2 more answers

7 lbs of sour candies cost $7.00. What is the unit price?

grandymaker [24]

Answer:

$1 per lb

Step-by-step explanation:

If you divide the cost by the amount you are getting, then you get the unit rate. So, since 7 divided by 7.00 is 1.00, then the cost per pound is $1.00

7 0

2 years ago

Judy Clark went to Reel Bank. She borrowed $7,800 at a rate of 6 1/2%. The date of the loan was September 2. Judy hoped to repay

abruzzese [7]

Answer:

$7995.85

Step-by-step explanation:

We will use simple interest formula to solve our given problem.

$A=P(1+rt)$ , where,

A = Amount after t years,

P = Principal amount,

r = Annual interest rate in decimal form,

t = Time in years.

$r=6.5\%=\frac{6.5}{100}=0.065$

$t=\text{141 days}=\frac{141}{365}\text{ year}$

$A=\$78001+0.065\times \frac{141}{365})$

$A=\$7800(1+0.065\times 0.38630136986)$

$A=\$7800(1+0.025109589041)$

$A=\$7800(1.025109589041)$

$A=\$7995.85479$

$A\approx \$7995.85$

Therefore, Judy will will pay back on January 20: $7995.85.

8 0

3 years ago

Suppose 52% of the population has a college degree. If a random sample of size 563563 is selected, what is the probability that

amm1812

Answer:

The value is $P(| \^ p - p| < 0.05 ) = 0.9822$

Step-by-step explanation:

From the question we are told that

The population proportion is $p = 0.52$

The sample size is n = 563

Generally the population mean of the sampling distribution is mathematically represented as

$\mu_{x} = p = 0.52$

Generally the standard deviation of the sampling distribution is mathematically evaluated as

$\sigma = \sqrt{\frac{ p(1- p)}{n} }$

=> $\sigma = \sqrt{\frac{ 0.52 (1- 0.52 )}{563} }$

=> $\sigma = 0.02106$

Generally the probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as

$P(| \^ p - p| < 0.05 ) = P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 ))$

Here $\^ p$ is the sample proportion of persons with a college degree.

$P( - (0.05 - 0.52 ) < \^ p < (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma } < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )$