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3 years ago

Given that x = 5.4 m and 0 = 26°, work out BC rounded to 3 SF.

Mathematics

1 answer:

bagirrra123 [75]3 years ago

4 0

Answer:

BC ≈ 4.85 m

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos26° = $\frac{adjacent}{hypotenuse}$ = $\frac{BC}{AC}$ = $\frac{BC}{5.4}$ ( cross- multiply )

5.4 × cos26° = BC , then

BC ≈ 4.85 m ( to 3 s f )

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1. Find the product. Simplify. 13/14 of 10/9. A. 15/42 B. 5/21 C. 2/7 D. 5/6 2. Andy has 5 pots, and each pot can hold 3/8 pound

Rufina [12.5K]

1) Resultant fraction: $\frac{65}{63}$

2) Total soil: $1\frac{7}{8}$ pounds

3) Result of the product: $\frac{35}{2}$

Step-by-step explanation:

In this problem, we want to find the product between the two fractions

$\frac{13}{14}$

and

$\frac{10}{9}$

In order to find this product, we have to multiply the numerators of each fraction and the denominators of each fraction. We get:

$\frac{13}{14}\cdot \frac{10}{9}=\frac{13\cdot 10}{14\cdot 9}=\frac{130}{126}$

Now we simplify, dividing both numerator and denominator by 2:

$\frac{130/2}{126/2}=\frac{65}{63}$

And the fraction cannot be further simplified.

Here we have:

n = 5 (number of pots that Andy has)

$s=\frac{3}{8}$ (amount of soil (in pounds) that each pot can contain)

In order to find the amount of soil that Andy needs to fill all the pots, we have to multiply the number of pots (n) by the amount of soil that each pot can contain (s).

If we do so, we find:

$t=n \cdot s = 5 \cdot \frac{3}{8}=\frac{5\cdot 3}{8}=\frac{15}{8}$

Which can be rewritten as a mixed fraction as:

$\frac{15}{8}=\frac{8+7}{8}=\frac{8}{8}+\frac{7}{8}=1\frac{7}{8}$

Here we want to find the result of the following product:

$4\frac{2}{3}\cdot 3\frac{3}{4}$

In order to do so, we first have to rewrite each fraction as an improper fraction.

For the 1st fraction:

$4\frac{2}{3}=\frac{4\cdot 3+2}{3}=\frac{12+2}{3}=\frac{14}{3}$

For the 2nd fraction:

$3\frac{3}{4}=\frac{3\cdot 4+3}{4}=\frac{12+3}{4}=\frac{15}{4}$

Now we can finally find the product of the two fractions:

$\frac{14}{3}\cdot \frac{15}{4}=\frac{14\cdot 15}{3\cdot 4}=\frac{210}{12}=\frac{35}{2}$

Where in the last step, we divide both the numerator and the denominator by 6.

Learn more about fractions:

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3 years ago

In gym class students were asked to form six equal groups. If there were 18 students in each group, then how many total students

Contact [7]

Answer:108

Step-by-step explanation:

18 times 6 is 108

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2 years ago

Country Financial, a financial services company, uses surveys of adults age and older to determine if personal financial fitness

Yakvenalex [24]

Complete Question

Country financials, a financial services company, uses surveys of adults age 18 and older to determine if personal financial fitness is changing over time. In February 2012, a sample of 1000 adults showed 410 indicating that their financial security was more that fair. In Feb 2010, a sample of 900 adults showed 315 indicating that their financial security was more than fair.

State the hypotheses that can be used to test for a significant difference between the population proportions for the two years.

What is the sample proportion indicating that their financial security was more that fair in 2012?In 2010?

Conduct the hypothesis test and compute the p-value.At a .05 level of significance what is your conclusion?

Answer:

The null hypothesis is $H_o : p_1 = p_2$

The alternative hypothesis is $H_a : p_1 \ne p_2$

in 2012 $\r p_1 =0.41$

in 2010 $\r p_2 =0.35$

The p-value is $p-value = 0.0072$

The conclusion is

There is sufficient evidence to conclude that the proportion of those indicating that financial security is more fair in Feb 2010 is different from the proportion of those indicating that financial security is more fair in Feb 2012.

Step-by-step explanation:

From the question we are told that

The sample size in 2012 is $n_1 = 1000$

The number that indicated that their finance was more than fail is k = 410

The sample size in 2010 is $n_2 = 900$

The number that indicated that their finance was more than fail is u = 315

The level of significance is $\alpha = 0.05[/ex]The null hypothesis is [tex]H_o : p_1 = p_2$

The alternative hypothesis is $H_a : p_1 \ne p_2$

Generally the sample proportion for 2012 is mathematically represented as

$\r p_1 = \frac{k}{n_1}$

=> $\r p_1 = \frac{410}{1000}$

=> $\r p_1 =0.41$

Generally the sample proportion for 2010 is mathematically represented as

$\r p_2 = \frac{u}{n_2}$

=> $\r p_2 = \frac{315}{900}$

=> $\r p_2 =0.35$

Generally the pooled sample proportion is mathematically represented as

$\r p = \frac{k + u }{n_1 + n_2}$

=> $\r p = \frac{410 + 315 }{1000+ 900}$

=> $\r p = 0.3816$

Generally the test statistics is mathematically represented as

$z = \frac{( \r p_1 - \r p_2 ) - 0}{\sqrt{\r p (1 - \r p ) ( \frac{1}{n_1} +\frac{1}{n_2} )} }$

$z = \frac{( 0.41 -0.35 ) - 0}{\sqrt{0.3816 (1 - 0.3816 ) ( \frac{1}{1000} +\frac{1}{900} )} }$

$z = 2.688$

Generally the p- value is mathematically represented as

$p-value = 2 P(Z >2.688 )$

From the z -table

$P(Z > 2.688) = 0.0036$

$p-value = 2 *0.0036$

$p-value = 0.0072$

So from the p-value obtained we see that p-value < $\alpha$ so we reject the null hypothesis

Thus there is sufficient evidence to conclude that the proportion of financial security is more fair in Feb 2010 is different from the proportion of financial security is more fair in Feb 2012.

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