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3 years ago

According to the triangle proportionality theorem, which proportion is correct?

Mathematics

2 answers:

butalik [34]3 years ago

7 0

Answer

MB/AM=NC/AN

hope it helps

vfiekz [6]3 years ago

3 0

According to triangle proportionality theorem:

If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.

And in the given triangle, MN parallel to BC, So MN divided the two sides in proportion, that is

$\frac{MB}{AM}=\frac{NC}{AN}$

Correct option is the third option .

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3 years ago

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3 years ago

Read 2 more answers

If the inspection division of a county weights and measures department wants to estimate the mean amount of soft-drink fill in 2

nikklg [1K]

Answer:

n=97

Step-by-step explanation:

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".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=0.01)$

We know that the margin of error for a confidence interval is given by:

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The next step would be find the value of $\z_{\alpha/2}$ , $\alpha=1-0.95=.05$ and $\alpha/2=0.025$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=1.96$

If we solve for n from formula (1) we got:

$\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}$

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And we have everything to replace into the formula:

$n=(\frac{1.96(0.05)}{0.01})^2 =96.04$

And if we round up the answer we see that the value of n to ensure the margin of error required $\pm=0.01 Liters$ is n=97.

5 0

4 years ago

a school netball team won ten matches , lost four and drew 1/8 of the total played . how many matches were drawn and what fracti

defon

The school net ball team drew 2 matches. The team won $\frac{5}{8}$ of total matches played

<h3>Solution:</h3>

Given, A school netball team won ten matches, lost four and drew $\frac{1}{8}$ of the total played

Let the total number of matches played be "n"

Number of matches drawn = $\frac{1}{8}$ of n

Total matches played = number of won matches + number of lost matches + number of drawn matches

$\begin{array}{l}{n=10+4+\frac{1}{8} n} \\\\ {n-\frac{1}{8} n=10+4}\end{array}$

Taking "n" as common from left hand side,

$\begin{array}{l}{\mathrm{n}\left(1-\frac{1}{8}\right)=14} \\\\ {\mathrm{n} \times \frac{8-1}{8} \quad 14} \\\\ {\mathrm{n} \times \frac{7}{8}=14} \\\\ {\mathrm{n}=16}\end{array}$

So, they played 16 matches in total

Number of matches drawn = $\frac{1}{8}$ of n = $\frac{1}{8} \times 16 = 2$ $\text { Fraction of won matches }=\frac{\text { number of matches won }}{\text { number of matches played }}=\frac{10}{16}=\frac{5}{8}$

$\text { Hence, the team drew } 2 \text { matches and won } \frac{5}{8} \text { of total played matches. }$

3 0

3 years ago