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olganol [36]
3 years ago
12

A navigator marks three locations on a satellite map as shown. If the distance from Serravalle to San Marino is 982 miles, and t

he distance from Chiesanuova to San Marino is 833 miles, then how far is Chiesanuova from Serravalle?
Mathematics
2 answers:
jeyben [28]3 years ago
6 0

Answer:

C .1,233 MILES

PLATO

Step-by-step explanation:

Solnce55 [7]3 years ago
3 0

                                                                    833
Seravalle____________Chiesanuova______________San Marino
    
             |---------------------------------------------------------------------|
                                          982

so the distance from Chiesanuova to Serravalle is : 
982 - 833 = 149 miles <==
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If h(x) = 5 x and k(x)=1/x, which expression is equivalent to (koh)(x)?
Lyrx [107]
May be there is an operator missing in the first function, h(x). I will solve this in two ways, 1) as if the h(x) = 5x and 2) as if h(x) = 5 + x

1) If h(x) = 5x and k(x) = 1/x

Then (k o h) (x) = k ( h(x) ) = k(5x) = 1/(5x)

2) If h(x) = 5 + x and k (x) = 1/x

Then (k o h)(x) =k ( h(x) ) = k (5+x) =  1 / [5 + x]
4 0
3 years ago
Read 2 more answers
K(x) = 6x+100 k(-5) =
kkurt [141]

Answer:

k(-5)=70

Step-by-step explanation:

The given function is

k(x)=6x+100

To find k(-5), we put x=-5 into the given function to get;

k(-5)=6(-5)+100

Multiply out to obtain;

k(-5)=-30+100

This is the same as;

k(-5)=100-30

This simplifies to

k(-5)=70

3 0
3 years ago
1/3 (12 - 24v) = -2(4v -2)
ANEK [815]
Opening brackets gives;
4-8v=-8v+4
Putting like terms together;
4-4=-8v+8v
0=0
There is no solution thus.
3 0
3 years ago
A process produces batches of a chemical whose impurity concentrations follow a normal distribution with a variance of175. A ran
Tom [10]

Answer:

the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)

Step-by-step explanation:

since the variance  S² of the batch follows a normal distribution , then for a sample n of  20 distributions , then the random variable Z:

Z= S²*(n-1)/σ²

follows a χ² ( chi-squared) distribution with (n-1) degrees of freedom

since

S² > 3.10 , σ²= 1.75 , n= 20

thus

Z > 33.65

then from χ² distribution tables:

P(Z > 33.65) = 0.02020

therefore the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)

8 0
3 years ago
Read 2 more answers
A skier has decided that on each trip down a slope, she will do 3 more jumps than before. On her first trip she did 5 jumps. Der
taurus [48]
Since we are already given the amount of jumps from the first trial, and how much it should be increased by on each succeeding trial, we can already solve for the amount of jumps from the first through tenth trials. Starting from 5 and adding 3 each time, we get: 5 8 (11) 14 17 20 23 26 29 32, with 11 being the third trial.

Having been provided 2 different sigma notations, which I assume are choices to the question, we can substitute the initial value to see if it does match the result of the 3rd trial which we obtained by manual adding.

Let us try it below:

Sigma notation 1:

  10
<span>   Σ (2i + 3)
</span>i = 3

@ i = 3

2(3) + 3
12

The first sigma notation does not have the same result, so we move on to the next.

  10
<span>   Σ (3i + 2)
</span><span>i = 3
</span>
When i = 3; <span>3(3) + 2 = 11. (OK)
</span>
Since the 3rd trial is a match, we test it with the other values for the 4th through 10th trials.

When i = 4; <span>3(4) + 2 = 14. (OK)
</span>When i = 5; <span>3(5) + 2 = 17. (OK)
</span>When i = 6; <span>3(6) + 2 = 20. (OK)
</span>When i = 7; 3(7) + 2 = 23. (OK)
When i = 8; <span>3(8) + 2 = 26. (OK)
</span>When i = 9; <span>3(9) + 2 = 29. (OK)
</span>When i = 10; <span>3(10) + 2 = 32. (OK)

Adding the results from her 3rd through 10th trials: </span><span>11 + 14 + 17 + 20 + 23 + 26 + 29 + 32 = 172.
</span>
Therefore, the total jumps she had made from her third to tenth trips is 172.


3 0
3 years ago
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