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Sav [38]
2 years ago
6

(7p^(2)+3p)-(5p^(2)+4) solve for standard form and show how you got it

Mathematics
1 answer:
inna [77]2 years ago
5 0

\huge \boxed{\mathfrak{Question} \downarrow}

  • ( 7 p ^ { 2 } + 3 p ) - ( 5 p ^ { 2 } + 4 ). Solve for standard form.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

( 7 p ^ { 2 } + 3 p ) - ( 5 p ^ { 2 } + 4 )

To find the opposite of 5p²+4, find the opposite of each term inside the bracket.

7p^{2}+3p-5p^{2}-4

Combine 7p² and -5p² to get 2p².

\boxed{ \boxed{ \bf \: 2p^{2}+3p-4 }}

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What are the steps to answer 28=8b+13b-35
oksian1 [2.3K]

Let's solve this problem step by step.

28=8b+13b-35

Step 1: Bring 35 to 28.

28=8b+13b-35

+35            +35

63=8b+13b

Step 2: Add 8b and 13b.

63=21b

Step 3: Divide both sides by 21.

63/21=21b/21

So, the answer for this problem is 3=b.

6 0
3 years ago
Please help i can’t figure this out
UNO [17]

Answer:

C because 100 percent of 15 is 15 and 20 percent of 15 is 3

6 0
2 years ago
Describe a process you would use to create the perpendicular bisector to a segment AB using only an unmarked straightedge and an
Murrr4er [49]

You'll have to c<span>ompass tip on A and draw a small ark with pencil approximately in the middle above AB line, now compass tip to point B and cross the ark you made previously.
Do the same on the opposite side without making any change to the compass 
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8 0
2 years ago
×-5y=-15<br> Complete the missing value in the solution to the equation<br> (-5,__)
Readme [11.4K]

Answer:

(-5  , 2 )

Step-by-step explanation:

hello :

×-5y=-15

5y = x+15

y = (1/5)x+3    for x= - 5    y = (1/5)(-5) +3  so :  y = 2

3 0
3 years ago
Two coins, A and B, each have a side for heads and a side for tails. When coin A is tossed, the probability it will land tails-s
BlackZzzverrR [31]

Answer:

Is the number of tosses for each coin enough for the sampling distribution of the difference in sample proportions PA-PB to be approximately normal?

b. No, 20 tosses for coin A is enough, but 20 tosses for coin B is not enough.

Step-by-step explanation:

a) Data and Calculations:

The probability of coin A landing tails-side up = 0.5

The proportion of times coin A lands tails-side up (PA) = 20 * 0.5 = 10

Therefore, the probability of coin A landing heads-side up = 0.5 (1 - 0.5)  

And the proportion of times that coin A lands heads-side up = 20 * 0.5 = 10.  

The proportion on either side is equally distributed.

This is why 20 tosses for coin A is enough, since the sample proportions PA is approximately normal, symmetric, and equally distributed.  There will be equal amounts of 10 tosses (0.5 *20) for either heads-side up or tails-side up.

For coin B, the probability of landing tails-side up = 0.8

The proportion of times coin B lands tails-side up (PB) = 20 * 0.8 = 16

Therefore, the probability of coin B landing heads-side up = 0.2 (1 - 0-.8)

The proportion on either side is not equally distributed, but skewed.

This is why 20 tosses for coin B is not enough, since the sample proportions PB is not approximately normal, symmetric, and equally distributed.  There will be 16 tosses landing tails-side up (0.8*20) and only 4 tosses landing heads-side up (0.2*20).

6 0
3 years ago
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