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krek1111 [17]
2 years ago
8

HELP ME, SOMEONE!!!

Mathematics
1 answer:
olya-2409 [2.1K]2 years ago
7 0

Answer:

10KL/1000L

8KL/800L

800L/8KL

Step-by-step explanation:

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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

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

<em />

<u>So, 99% confidence interval for the true mean yield, </u>\mu<u> is ;</u>

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

<u>80% confidence interval for</u> \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
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