explain how you can use multiplication to find the quotient 3/5 divided by 3/15. then evaluate the expression

Mathematics

2 answers:

Mariana [72]2 years ago

8 0

we know that

Dividing is the opposite of multiplying. Therefore, dividing by a fraction can be accomplished by multiplying by its reciprocal

$\frac{\frac{a}{b}}{\frac{c}{d}} =\frac{a}{b}*\frac{d}{c}$

in this problem we have

$\frac{a}{b}=\frac{3}{5}\\ \\\frac{c}{d}=\frac{3}{15}$

substitute in the formula above

$\frac{\frac{3}{5}}{\frac{3}{15}} =\frac{3}{5}*\frac{15}{3}=3$

therefore

<u>the answer is </u>

$3$

Vaselesa [24]2 years ago

7 0

$\frac{3}{5}$ ÷ $\frac{3}{15}$

⇒ $\frac{3}{5}$ × $\frac{15}{3} = \frac{45}{15} = 3$

You might be interested in

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05