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3 years ago

What is the approximate distance between the points (-3,-4)(-8,1) on a coordinate grid?

1 answer:

Alex3 years ago

3 0

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-3}~,~\stackrel{y_1}{-4})\qquad (\stackrel{x_2}{-8}~,~\stackrel{y_2}{1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[-8-(-3)]^2+[1-(-4)]^2}\implies d=\sqrt{(-8+3)^2+(1+4)^2} \\\\\\ d=\sqrt{25+25}\implies d=\sqrt{50}\implies d\approx 7.071$

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3 years ago

Can someone help please

vova2212 [387]

Answer:

24sq.ft

Step-by-step explanation:

<h3>Area of a parallelogram = Base*Height</h3>

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→ 6*4ft

→24sq.ft

<h3 /><h3>While writing area of a figure, we have to express the area in sq which means square</h3>

<h2> <em>ThankYou</em></h2>

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2 years ago

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Virty [35]

Answer:

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3 years ago

Expressed as a product of its prime factors in index form, a number N is

Elena-2011 [213]

<u>ANSWER</u>

$5N^2=3^{2} \times 5^{5} \times x^{6}$

<u>EXPLANATION</u>

$N=3\times5^2 \times x^3$ .

$5N^2=5(3\times5^2 \times x^3)^2$

Recall this property of exponents;

$(a^m)^2=a^{m} \times a^m$

So our product becomes;

$5N^2=5(3\times5^2 \times x^3) \times (3\times5^2 \times x^3)$

$5N^2=5\times 3\times 3 \times 5^2 \times 5^2 \times x^3 \times x^3$

$5N^2=3\times 3\times 5 \times 5^2 \times 5^2 \times x^3 \times x^3$

Recall this law of exponents:

$a^m \times a^n =a ^{m+n}$

$5N^2=3^{1+1} \times 5^{1+2+2} \times x^{3+3}$

$5N^2=3^{2} \times 5^{5} \times x^{6}$

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3 years ago

Suppose that a company prints baseball cards. They claim that 30% of the cards feature rookies, 60% feature veterans, and 10% fe

liraira [26]

Answer:

H0: The distribution of players featured on the cards is 0.30 rookies, 0.60 veterans, and 0.10 All-Stars.

Ha: At least one of the proportions in the null hypothesis is false.

Step-by-step explanation:

On this case we need to apply a Chi squared goodness of fit test, and the correct system of hypothesis would be:

H0: The distribution of players featured on the cards is 0.30 rookies, 0.60 veterans, and 0.10 All-Stars.

Ha: At least one of the proportions in the null hypothesis is false.

And in order to test it we need to have observed and expected values. On this case we can calculate the Expected values like this

$E_{rookies}=50*0.3=15$

$E_{veterans}=50*0.6=30$

$E_{All stars}=50*0.1=5$

The observed values are not provided. The statistic on this case is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i) -E_i}{E_i}$

And this statistic follows a chi square distribution with k-1 degrees of freedom on this case k=3, since we have 3 groups.

We can calculate the p valu like this:

$P(\chi^2 > \chi^2_{calculated})=p_v$

And if the p value it's higher than the significance level we FAIL to reject the null hypothesis. In other case we reject the null hypothesis.

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3 years ago