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3 years ago

What is the greatest common factor GCF of 10 and four

Mathematics

2 answers:

Feliz [49]3 years ago

7 0

Answer:

Step-by-step explanation:

thats the biggest number that goes with both

Pani-rosa [81]3 years ago

3 0

Answer:

Take the numbers 50 and 30. Their greatest common factor is 10, since 10 is the greatest factor that both numbers have in common. To find the GCF of greater numbers, you can factor each number to find their prime factors, identify the prime factors they have in common, and then multiply those together.

Step-by-step explanation:

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The net of a triangular prism and its dimensions are shown in the diagram. what is the lateral surface area of the triangular pr

Furkat [3]

Answer: 120 sq in.

Step-by-step explanation:

5 0

3 years ago

[Geometry Basics] I'm just checking if this is correct:

Leno4ka [110]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ x &,& y~) 
% (c,d)
&&(~ -9 &,& -1~)
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right)
\\\\\\
\left( \cfrac{-9+x}{2}~~,~~\cfrac{-1+y}{2} \right)=\stackrel{midpoint}{(8,14)}\implies 
\begin{cases}
\cfrac{-9+x}{2}=8\\\\
-9+x=16\\
\boxed{x=25}\\
-------\\
\cfrac{-1+y}{2} =14\\\\
-1+y=28\\
\boxed{y=29}
\end{cases}$

$\bf -------------------------------\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ x &,& y~) 
% (c,d)
&&(~10 &,& 12~)
\end{array}\qquad
\\\\\\
\left( \cfrac{10+x}{2}~~,~~\cfrac{12+y}{2} \right)=\stackrel{midpoint}{(6,9)}\implies 
\begin{cases}
\cfrac{10+x}{2}=6\\\\
10+x=12\\
\boxed{x=2}\\
-------\\
\cfrac{12+y}{2} =9\\\\
12+y=18\\
\boxed{y=6}
\end{cases}$

3 0

3 years ago

Scores on the SAT Mathematics test are believed to be normally distributed. The scores of a simple random sample of five student

AysviL [449]

Answer:

The mean calculated for this case is $\bar X=584$

And the 95% confidence interval is given by:

$584-2.776\frac{86.776}{\sqrt{5}}=476.271$

$584+2.776\frac{86.776}{\sqrt{5}}=691.729$

So on this case the 95% confidence interval would be given by (476.271;691.729)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=584$

The sample deviation calculated $s=86.776$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=5-1=4$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,4)".And we see that $t_{\alpha/2}=2.776$