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2 years ago

61 is the sum of 18 and vidyas age

Mathematics

1 answer:

liq [111]2 years ago

3 0

To find to find this answer all you need to do is subtract 18 from 61.

61-18=43

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Given the data set (200,150,325,100,75) find the mean of the data set

Nina [5.8K]

To find the mean (average), u add up all the numbers, then divide by how many numbers there are.

(200 + 150 + 325 + 100 + 75) / 5 = 850/5 = 170 <== the mean

7 0

2 years ago

If your purchases of shoes increase from 9 pairs per year to 11 pairs per year when your income increases from $19,000 to $21,00

Leni [432]

Answer: commom resources good

Step-by-step explanation:

The question above signifies that I can't eliminate shoes from my list. To me, shoes are inexcludable. You can see that as my income increases, my purchases of shoes increase.

I chose common resources good because they are non-excludable (something difficult to eliminate), but can compete with other goods. This means that their availability to other people decreases when they are being used up.

Other types of goods besides common resources good are

- Private goods (products that are easily eliminated and rival)

-Club goods (excludable products but non rival)

-Public goods (unable to eliminate and non-rival)

8 0

2 years ago

Help meh with this lol

egoroff_w [7]

Answer:

Step-by-step explanation:

Integers: a WHOLE number that can be positive, negative or zero

-22 is a negative whole number

Rational number: Any number that can be written as a fraction with integers.

-22/1 can be the fraction

3 0

3 years ago

Read 2 more answers

Sofia has four times as many dollars as her friend jessie has. together the two have $35. how much money do they each have?

garik1379 [7]

4J = s
J + s = J + 4j = 5j = 35

J = 7
s = 28

3 0

2 years ago

Show all work and reasoning

Natalija [7]

Split up the interval [2, 5] into $n$ equally spaced subintervals, then consider the value of $f(x)$ at the right endpoint of each subinterval.

The length of the interval is $5-2=3$ , so the length of each subinterval would be $\dfrac3n$ . This means the first rectangle's height would be taken to be $x^2$ when $x=2+\dfrac3n$ , so that the height is $\left(2+\dfrac3n\right)^2$ , and its base would have length $\dfrac{3k}n$ . So the area under $x^2$ over the first subinterval is $\left(2+\dfrac3n\right)^2\dfrac3n$ .

Continuing in this fashion, the area under $x^2$ over the $k$ th subinterval is approximated by $\left(2+\dfrac{3k}n\right)^2\dfrac{3k}n$ , and so the Riemann approximation to the definite integral is

$\displaystyle\sum_{k=1}^n\left(2+\frac{3k}n\right)^2\frac{3k}n$

and its value is given exactly by taking $n\to\infty$ . So the answer is D (and the value of the integral is exactly 39).

8 0

2 years ago