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3 years ago

14

in her last basketball game, Chloe scored 8 points. Today, she had a personal best and score 15 points. Calculate the percent ch

ange

1 answer:

AysviL [449]3 years ago

7 0

Answer:

46.666 (repeating 6) or 46 and 2/3

Step-by-step explanation:

1. Find the difference, 15 - 8 = 7.

2. Divide, 7 / 15 = 0.466..

3. Change to percent. (Multiply)

(0.466... x 100) = 46.66..

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prove that if f is integrable on [a,b] and c is an element of [a,b], then changing the value of f at c does not change the fact

Neko [114]

Answer with Step-by-step explanation:

We are given that if f is integrable on [a,b].

c is an element which lie in the interval [a,b]

We have to prove that when we change the value of f at c then the value of f does not change on interval [a,b].

We know that limit property of an integral

$\int_{a}^{b}f dt=\int_{a}^{c}fdt+\int_{c}^{b} fdt$

$\int_{a}^{b} fdt=f(b)-f(a)$ ....(Equation I)

Using above property of integral then we get

$\int_{a}^{b}fdt=\int_{a}^{c}fdt+\int_{c}^{b} fdt$ ......(Equation II)

Substitute equation I and equation II are equal

Then we get

$\int_{a}^{b}fdt= f(c)-f(a)+{f(b)-f(c)}$

$\int_{a}^{b}fdt=f(c)-f(a)+f(b)-f(c)=f(b)-f(a)$

$\int_{a}^{c}fdt+\int_{c}^{b}fdt=f(b)-f(a)$

Therefore, $\int_{a}^{b}fdt=\int_{a}^{c}fdt+\int_{c}^{b}fdt$ .

Hence, the value of function does not change after changing the value of function at c.

6 0

3 years ago

Quinton answered 90% of his test questions correctly. Quinton answered 54 questions correctly.

snow_tiger [21]

There would be 60 questions on the test.

90%=54
1%= 54/90 = 6/10
100%= 6×100/10 =6×10=60

7 0

3 years ago

An auto parts store is examining how many items were purchased per transaction at the store as compared to their online website.

anastassius [24]

Answer:

0.4394

Step-by-step explanation:

given that an auto parts store is examining how many items were purchased per transaction at the store as compared to their online website

The data is shown below:

Frequency Frequency*Midpt

Midpt Online Total Online Total

1-3 2 40 147 80 294

4-6 5 60 103 300 515

7-9 8 15 32 120 256

10-12 11 5 18 55 198

120 300 555 1263

Thus we find total items purchased on line = 555

and total items overall purchased = 1263

Hence the probability that a randomly selected item was purchased online

= $\frac{555}{1263} \\=0.4394$

4 0

3 years ago

Please help me with this one I beg you!! And don’t send those file links I know that their scams. Please help I will mark brainl

borishaifa [10]

Step-by-step explanation:

2;6 -> 6; -2

4;4 -> 4 ; -4

6;8 -> 8; -6

sorry English isn't my first language so I can't explain it :((

3 0

3 years ago

I need help with the question please!

Elanso [62]

2x + 4(x - 1) = 2 + 4x
2x + 4x - 4 = 2 + 4x
6x - 4 = 2 + 4x
6x - 4x = 2 + 4
2x = 6
x = 6/2 = 3
x = 3
=> x has one solution.

3 0

4 years ago