HELP ME, SOMEONE!!!

A truck with a heavy load drove from Boston to New York at 50 mph. After dropping of the load, it returned from New York to Bost

leva [86]

Answer:

175/3 or 58.333...

Step-by-step explanation:

This is fairly complicated but I will try to make it as simple as possible. I also apologize in advance for how impossible it is to make fractions look like fractions. I also had to insert a bunch of unnecessary parentheses just because the fractions that I can make are relatively inaccurate.

The distance between 2 cities can be represented as d.

Time is d/r if r is rate, so B->NY=d/50

Similarly,NY->B is d/70

Obviously the average speed is 2d/(d/50+d/70), this is just an average time

Now just remove the d and get (d/d)*2/(1/50+1/70)

You can now multiply by the LCM/LCM, which is 350/350.

After calculating you will get 700/12 which is 58.333...(This is how to enter into your RSM browser) Hope this helps. Sorry it is so late.

8 0

3 years ago

Mario needs 32.5 m of chain link fence to enclose his garden. The cost of the fence is $6.88 per meter.

DaniilM [7]

Cost of fence for a meter = 6.88

For 32.5 m, it would be = 6.88 * 32.5 = 223.6

D)223.60 is answer

8 0

3 years ago

Read 2 more answers

Which expressions are equivalent to the one below check that all apply 25^x/5^x

madreJ [45]

C and D apply to your expression.

7 0

2 years ago

Read 2 more answers

8. rajis annual salary ranges from $25 325 in the 1st year to $34 445 in the 7th year. The salaries in this range form an arithm

-Dominant- [34]

Answer:

a) $1520

b) $S_7 =$ $209195

Step-by-step explanation:

The range of salaries forms an arithmetic sequence with the first term as 25325 and its 7th term as 34445:

a) The raise the person can get each year is the common difference of the progression, d.

The nth term of an arithmetic progression is given generally as:

$a_n = a + (n - 1)d$

where a = first term = 25325

d = common difference

Therefore, the 7th term (34445) will be:

34445 = 25325 + (7 - 1)d

34445 = 25325 + 6d

=> 6d = 34445 - 25325 = 9120

d = 9120 / 6 = $1520

Therefore, the raise the person gets each year (common difference) is $1520.

b) The total amount the person will earn after 7 years is the sum of the salaries of all 7 yeas.

The sum of an arithmetic progression up to the nth term is given as:

$S_n = \frac{n}{2}(2a + (n - 1)d)\\ \\$

Therefore, the sum of the person's salary for the 7 years is:

$S_7 = \frac{7}{2}(2 * 25325 + (7 - 1)1520)\\ \\S_7 = 3.5 (50650 + 6(1520))\\\\S_7 = 3.5 (50650 + 9120)\\\\S_7 = 3.5 * 59770\\$

$S_7 =$ $209195

That is the total amount of salary after 7 years

5 0

3 years ago

A random sample of 9 fields of corn has a mean yield of 40.6 bushels per acre and standard deviation of 7.52 bushels per acre. D

zaharov [31]

Answer:

The critical value that should be used in constructing the confidence interval is -1.397 and 1.397.

80% confidence interval for the true mean yield is [37.1 bushels per acre, 44.1 bushels per acre].

Step-by-step explanation:

We are given that a sample of 1584 third graders, the mean words per minute read was 35.7. Assume a population standard deviation of 3.3.

Firstly, the pivotal quantity for 80% confidence interval for the true mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean yield = 40.6 bushels per acre

s = sample standard deviation = 7.52 bushels per acre

n = sample of fields of corn = 9

$\mu$ = true mean yield

Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about the population standard deviation.

So, 99% confidence interval for the true mean yield, $\mu$ is ;

P(-1.397 < $t_8$ < 1.397) = 0.80 {As the critical value of t at 8 degree of

of freedom are -1.397 & 1.397 with P = 10%}

P(-1.397 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.397) = 0.80

P( $-1.397 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.397 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.397 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.397 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.397 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.397 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $40.6-1.397 \times {\frac{7.52}{\sqrt{9} } }$ , $40.6+1.397 \times {\frac{7.52}{\sqrt{9} } }$ ]

= [37.1 , 44.1]

Therefore, 80% confidence interval for the true mean yield is [37.1 bushels per acre, 44.1 bushels per acre].

7 0

3 years ago