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3 years ago

Please help me, i will give brainiest !!

Mathematics

2 answers:

BARSIC [14]3 years ago

8 0

Answer:

If Cora runs 1/2 mile per week and x is the number of weeks, then she will run 5+1/2x. Then determine how many weeks are in two and a half months (depends on which month it is as there are less weeks in February than in January!) but we could estimate 6.

For Hana, you will have 3+ 3/4 x

Step-by-step explanation:

First equation is y=5+1/2 x and then plug 6 in for x to solve for y. the second equation is y=3+3/4x and plug in 6.

svet-max [94.6K]3 years ago

3 0

Answer:

If Cora runs 1/2 mile per week and x is the number of weeks, then she will run 5+1/2x.

Step-by-step explanation:

basically this explains the context of the problem

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Brandon bought 21 chicken wings for $39.90. How much would it cost for 15 wings?

Pachacha [2.7K]

Answer:

$28.50

Step-by-step explanation:

21 chicken wings= $39.90

1 chicken wing= 39.90÷21= $1.90

15 chicken wings= 15×1.90= $28.50

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3 years ago

Rectangle and a square have the same area. the length of the rectangle is seventy feet more than two times its width. the length

Natasha2012 [34]

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3 years ago

The volume of a rectangular box is (x3 - 7x2 - 9x + 63) cubic units. Determine the dimensions of the rectangular box by factorin

Shalnov [3]

Hello,
well done.
As x^3-7x²-9x+63=(x-3)(x+3)(x-7) we must have x>3 and x>-3 and x>7 thus x>7

Explanations:
x^3-7x²-9x+63=x^3-9x+(-7x²+63)=x(x²-9)-7(x²-9)=(x²-9)(x-7)=(x+3)(x-3)(x-7)

6 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

5. Given the following triangle side lengths, identify the triangle as acute, right or obtuse. Show your work.

Roman55 [17]

Refer to the figure shown below.
We shall review each of the three given measurements and decide what type of triangle we have.

Measurement a.
a=3, b=4, c=5.
For a right triangle, c² = a² + b² (Pythagorean theorem)
a² + b² = 3² + 4² = 9 + 16 = 25
c² = 5² = 25
Answer:
This is a right triangle, because c² = a² + b².

Measurement b.
a=5, b=6, c=7.
For an acute triangle, c² < a² + b².
a² + b² = 5² + 6² = 25 + 36 = 61
c² = 7² = 49
Answer:
This is an acute triangle, because c² < a² + b².

Measurement c.
a=8, b=9, c=12.
For an obtuse triangle, c² > a² + b².
a² + b² = 8² + 9² = 64 + 81 = 145
c² = 12² = 144
Answer:
This is an acute triangle because c² < a² + b².

6 0

3 years ago

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