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4 years ago

I need help solving this :/

Mathematics

1 answer:

schepotkina [342]4 years ago

4 0

The equivalents are ...

A. 10 · (4 + 3)
B. 10 · 3 + 10 · 4
C. (3 + 4) · 10

_____

The expression in D evaluates to 30+4 = 34, not 10·7 = 70.

In A, (4 + 3) is equivalent to (3 + 4) by the commutative property of addition. The commutative properties of multiplication and addition allow you to swap the order of the operands with no effect on the result.

In B, 10·3 +10·4 is equivalent to 10·(3+4) by the distributive property of multiplication over addition.

In C, (3+4)·10 is equivalent to 10·(3+4) by the commutative property of multiplication.

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Simplify.<br> Rewrite the expression in the form 4^n

lys-0071 [83]

Answer:

4^(-2)

Step-by-step explanation:

4^6 * 4^-8

Since the bases are the same, and we are multiplying, we can add the exponents

4^(6-8)

4^(-2)

8 0

3 years ago

Read 2 more answers

The credit remaining on a phone card (in dollars) is a linear function of the total calling time made with the card (in minutes)

Ivan

Answer: the remaining credit after 79 minutes of call is $15.52

Step-by-step explanation:

let x represent the number of minutes of call.

Let y represent the remaining credit after x minutes of call.

If we plot y on the vertical axis and x on the horizontal axis, a straight line would be formed. The slope of the straight line would be

Slope, m = (17.2 - 19.84)/(65 - 43)

m = - 2.64/22 = - 0.12

The equation of the straight line can be represented in the slope-intercept form, y = mx + c

Where

c = intercept

m = slope

To determine the intercept, we would substitute x = 65, y = 17.2 and m = - 0.12 into y = mx + c. It becomes

17.2 = - 0.12 × 65 + c

17.2 = - 7.8 + c

c = 17.2 + 7.8

c = 25

The linear function becomes

y = - 0.12x + 25

Therefore, the remaining credit after 79 minutes of call is

y = - 0.12 × 79 + 25

y = - 9.4 + 25

y = 15.52

6 0

3 years ago

Write an expression to represent the area of a trapezoid which can be found my multiplying the height of the trapezoid by half o

erma4kov [3.2K]

2(a^2+b^2).

A^2 but b^2 = c^2 (area of triangle)

A trapezoid is two triangles put together i just say multiply those together

7 0

3 years ago

PLS HELP ASAP THANKS ILL GIVE BRAINLKEST PLS THANKS PLS ASAP PLS PLS

Ivan

Answer:

12.5 inches

Step-by-step explanation:

2 / 4 = 0.5

0.5 in = 10ft

0.5 x 25 = 12.5

8 0

3 years ago

Read 2 more answers

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

7 0

4 years ago