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3 years ago

Substitute (–5, 0.5) into x – 4y = –7 to get .

Mathematics

1 answer:

lesya [120]3 years ago

8 0

Answer:

-7

Step-by-step explanation:

Coordinates are listed as (x,y) so x = -5 and y = .5

If you substitute the numbers in place of the letters, you get:

(-5) - 4(.5) = -7

-5 - 2 = -7

-7 = -7

Marking as Brainliest is much appreciated.

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What is 150,000,000 in scientific notion, PLEASE answer asap im so far behind!!

dybincka [34]

To write a number in scientific notation, you need to write the number as a product of a number from 1 to under 10 multiplied by a integer power of 10.

First, what number from 1 to under 10 can you get out of the digits of 150,000,000 just by changing the decimal point? The answer is 1.5 since 1.5 is greater than or equal to 1 and less than 10.

150,000,000 = 1.5 * 100,000,000

Now we change 100,000,000 into a power of 10. A 1 followed by a number of zeros is the same as 10 to the power equal to the number of zeros. In 100,000,000 the 1 is followed by 8 zeros, so 100,000,000 = 10^8.

150,000,000 = 1.5 * 10^8

6 0

3 years ago

Omg, I need help! A builder is buying property where she can build new houses. The line plot shows the sizes for each house. 1/6

klemol [59]

Answer:

Average size of the lots = ⅓ acre

Step-by-step explanation:

The question incomplete without specifying what we are to determine.

Question:A builder is buying property where she can build new houses. The line plot shows the sizes for each house. 1/6 has 6 X's 1/3 has 3 X's and 1/2 has 6 X's. Organize the information in a line plot. What is the average size of the lots? _________ acre

Help anyone?

Solution:

We are asked to organize the information in a line plot. See attachment for the line plot.

Given: 1/6 has 6 X's 1/3 has 3 X's and 1/2 has 6 X'sIn no particular order, the sizes of the lots are:1/6, 1/6, 1/6, 1/6, 1/6, 1/6, 1/3, 1/3, 1/3, 1/2, 1/2, 1/2, 1/2, 1/2 and 1/2 acre.

Let's count the number of lot for each size given.

For 1/6: there are 6 X's on the line plot of 1/6 number of lot for 1/6

= the lot × number of times it occurs

= (1/6) × 6 = 6/6 = 1 acre

For 1/3: there are 3 X's on the line plot of 1/3 number of lot for 1/3 = the lot × number of times it occurs

= (1/3) × 3 = 3/3 = 1 acre

For 1/2: there are 6 X's on the line plot of 1/2 number of lot for 1/2

= the lot × number of times it occurs

= (1/2) × 6 = 6/2= 3 acres

To find average size of the lots, we would sum all lot for each given size then divide by the total number of lots given.Sum of all lot for each given size = 1+1+3

Sum of all lot for each given size = 5

The total number of lots given = 15

Average size of the lots = 5/15 = 1/3

Average size of the lots = ⅓ acre

7 0

3 years ago

Will give brainliest answer

vovangra [49]

Answer:

A=50.26548246units^2

Step-by-step explanation:

Radius is 4 because half the diameter (8) is the radius (4)

A=πr^2

A=π(4)^2

A=50.26548246units^2

8 0

3 years ago

Read 2 more answers

H(x)=x^2-4x+1 find axis of symmetry

Tresset [83]

Answer:

the answer is x=2

I used math-way

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4 years ago

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