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Lesechka [4]
3 years ago
14

We have 8 yards of wraping paper. if we use 2 feet for each present how many present can we wrap

Mathematics
2 answers:
Ber [7]3 years ago
6 0
You can wrap 12 presents if you have 8 yards of wrapping paper and use 2 feet for each present.
kirill [66]3 years ago
3 0
We can wrap 12 presents. multiply the number of yards by the feet.
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Suppose A and B are points. How many lines contain both A and B?
spayn [35]
One and only one line contains both A and B (assuming, that A and B do not overlap)
7 0
2 years ago
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A research report says that t(63) = 1.99; p = .03. From that information, can you reject the null hypothesis with 95% confidence
Leona [35]

Answer:

Yes, I can reject the null hypothesis with 95% confidence.

Step-by-step explanation:

Critical value t(63) = 1.99

p-value = 0.03

Confidence level (C) = 95% = 0.95

Significance level = 1 - C = 1 - 0.95 = 0.05

Conclusion:

Reject the null hypothesis because the p-value 0.03 is less than the significance level 0.05.

4 0
2 years ago
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A ratio is... A) the difference between two numbers B) a comparison of two numbers by division C) a comparison of two numbers by
nlexa [21]
C
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3 0
2 years ago
Can someone please help me! Big ideas 8.4 number 16
8_murik_8 [283]
<h3>Answer:   24.5 </h3>

============================================================

Explanation:

It might help to draw it out. See the diagram below.

The base is JK which is 7 units long. I'm picking this as the base because this segment is completely horizontal.

Notice how the base JK spans from x = -3 to x = 4, which we can use subtraction and absolute value to get |J-K| = |-3-4| = 7.

The height is always perpendicular to the base. If the base is JK = 7, then the height is the vertical distance from K (or J) to L. This vertical distance is also 7 units. Subtract the y coordinates and apply absolute value. The base and height aren't always the same number.

Now we can find the area of the triangle

area = base*height/2

area = 7*7/2

area = 49/2

area = 24.5 square units

3 0
2 years ago
a. How wide must a wall footing be if the load is 9,500 pounds per foot of wall length, and the footing rests on a sandy gravel?
Phantasy [73]

Answer:

a) 38"

b) L = 5.7'

Step-by-step explanation:

a) Given

P = 9,500 lb/ft

Thick = 18"

We assume an allowable bearing pressure of σ = 3000 lb/sf

If Stress = Force / Area

σ = P/A

Solving for area

A = P/σ  ⇒  A = 9,500 lb/ft/3000 lb/sf

⇒  A = 3.167 sf,  or  3.167 ft wide, per foot of length.

⇒  A = 38" wide

We can see the pic 1 in order to understand the answer.

b) Given

P = 65,000 pounds = 65,000 lb

Thick = 18" = 1.5'

We assume an allowable bearing pressure of σ = 2000 lb/sf

If Stress = Force / Area

σ = P/A

Solving for area

A = P/σ  ⇒  A = 65,000 lb/2000 lb/sf

⇒  A = 32.5 sf

then A = L²  ⇒  L = √A = √(32.5 sf) = 5.7 ft

Finally L = 5.7'

We can see the pic 2 in order to understand the answer.

3 0
2 years ago
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