All categories

2 years ago

How do you write 0.12 as a fraction

Mathematics

1 answer:

Arturiano [62]2 years ago

7 0

[ Answer ]

$\boxed{\frac{3}{25} }$

[ Explanation ]

Rewrite 0.12 As A Fraction

-----------------------------

Rewrite Decimal As A Fraction With 1 As A Denominator

0.12 = $\frac{0.12}{1}$

Multiply To Remove Decimal Places

$\frac{0.12}{1}$ · $\frac{100}{100}$ = $\frac{12}{100}$

Find GCF, Reduce Fraction

$\frac{12 \ / \ 4}{100 \ / \ 4}$ = $\frac{3}{25}$

0.12 = $\frac{3}{25}$

$\boxed{\boxed{[ \ Eclipsed \ ]}}$

You might be interested in

Find the slope of the line that passes through (6, 68) and (54,-7)

soldi70 [24.7K]

Answer:

-25/16

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(-7-68)/(54-6)

m=-75/48

m=-25/16

7 0

1 year ago

Read 2 more answers

Given f (x) =2x-3 what is f(-15) A. -16 B. -13 C. 7 D. 30

Murljashka [212]

Letter d is the answer.

5 0

2 years ago

The table on the right shows last initials of basketball players and the number of games played by each. Find the z-score for p

Umnica [9.8K]

The answer is J 73
a a

5 0

3 years ago

A cylinder has a base area of 100cm and volume of 1500 cm what’s the height

e-lub [12.9K]

Answer:

we know,

Volume of cylinder = Base area × height

1500= 100×h

so, Height h = 15 cm

6 0

1 year ago

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

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

2 years ago