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Lyrx [107]
3 years ago
5

How lond did the famouse war of 100 years last ?

Mathematics
1 answer:
tatyana61 [14]3 years ago
6 0

Answer:

It lasted 116 years and saw many major battles – from the battle of Crécy in 1346 to the battle of Agincourt in 1415, which was a major English victory over the French.

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Find the midpoint of the line segment with the given endpoints. (-7, -1) and (3, -6)
Alchen [17]

Answer:

(-2, -3.5)

The middle of -7 and 3 is -2. The middles of -1 and -6 is -3.5

Vote me brainliest :D

3 0
4 years ago
Which choices are equivalent to the expression below? Check all that apply.
nignag [31]
The answer to it is A 12x
8 0
3 years ago
How would the expression x^3+64 be written using sum of cubes
Hunter-Best [27]
\bf \textit{difference and sum of cubes}
\\\\
a^3+b^3 = (a+b)(a^2-ab+b^2)
\\\\
a^3-b^3 = (a-b)(a^2+ab+b^2)\\\\
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\boxed{64=4^3}\qquad \qquad x^3+64\implies x^3+4^3\implies (x+4)(x^2-4x+16)
4 0
3 years ago
Read 2 more answers
Farmers know that driving heavy equipment on wet soil compresses the soil and injures future crops. Here are data on the "penetr
Greeley [361]

Answer:

Step-by-step explanation:

Hello!

To see if driving heavy equipment on wet soil compresses it causing harm to future crops, the penetrability of two types of soil were measured:

Sample 1: Compressed soil

X₁: penetrability of a plot with compressed soil.

n₁= 20 plots

X[bar]₁= 2.90

S₁= 0.14

Sample 2: Intermediate soil

X₂: penetrability of a plot with intermediate soil.

n₂= 20 (with outlier) n₂= 19 plots (without outlier)

X[bar]₂= 3.34 (with outlier) X[bar]₂= 2.29 (without outlier)

S₂= 0.32 (with outlier) S₂= 0.24 (without outlier)

Outlier: 4.26

Assuming all conditions are met and ignoring the outlier in the second sample, you have to construct a 99% CI for the difference between the average penetration in the compressed soil and the intermediate soil. To do so, you have to use a t-statistic for two independent samples:

Parámeter of interest: μ₁-μ₂

Interval:

[(X[bar]₁-X[bar]₂)±t_{n_1+n_2-2;1-\alpha/2}*Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }]

t_{n_1+n_2-2;1-\alpha/2}= t_{20+19-2;1-(0.01/2)}= t_{37; 0.995}= 2.715

Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{19*0.0196+18*0.0576}{20+19-2} }= 0.195= 0.20

[(2.90-2.29)±2.715*0.20\sqrt{\frac{1}{20} +\frac{1}{19} }]

[0.436; 0.784]

I hope this helps!

5 0
3 years ago
5)<br> I<br> 6<br> 50°<br> 16x + 2<br> B) 9<br> A) -10<br> C) 4<br> D) 8
dalvyx [7]

Answer:

c is answer

Step-by-step explanation:

c is ans because it is simple c only serves the need of questions and satisfy it

8 0
3 years ago
Read 2 more answers
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