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3 years ago

Find the least common multiple to determine the least common denominator for 2/3, 5/6 and 9/4

2 answers:

6 0

Rewriting input as fractions if necessary:
2/3, 5/6, 9/4

For (3, 6, 4) the least common multiple (LCM) is 12.

Therefore, the least common denominator (LCD) is 12.

Rewriting as equivalent fractions with the LCD:
8/12, 10/12, 27/12

Elanso [62]3 years ago

6 0

Mutiply 3 by 4 Mutiply 6 by 2 Mutiply 4 by 3 Answer 12

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What is the least amount of sap any one tree produced

pashok25 [27]

On average, a tapped maple will produce 10 to 20 gallons of sap per tap.

1/4

This answer is attached to a image, which has options: 1/4, 3/8, 5/8, 1.

So, the option that express the least amount of sap is the first one.

-Rin

5 0

3 years ago

Read 2 more answers

What does 12A equal?

lozanna [386]

4a+6b=10
4a=10-6b
a=(10-6b)/4
a=(5-3b)/2

2a-4b=12
2((5-3b)/2)-4b=12
(5-3b)-4b=12
5-7b=12
-7b=7
b=-1

2a-4b=12
2a-4(-1)=12
2a+4=12
2a=8
a=4

a=4, b=-1 

12a=48

7 0

3 years ago

Read 2 more answers

Question 8 (5 points) 4(8+2*5-6) 164 ob 24 176 o od 48

Vilka [71]

Answer:

Step-by-step explanation:

4(8+2(5)-6)

=4(8+10-6)

=4(12)

=48

5 0

4 years ago

The mean water temperature downstream from a power plant coolingtower discharge pipe should be no more than 102oF. Pastexperienc

nata0808 [166]

Answer:

(a) Yes, there is evidence that the water temperature is acceptable at $\alpha = 0.05$ (b) 0.9987 (c) 6.647274e-06

Step-by-step explanation:

Let X be the random variable that represents the water temperature. The water temperature has been measured on n = 9 randomly chosen days (a small sample), the sample average temperature is $\bar{x}$ = 100°F and $\sigma$ = 2°F. We suppose that X is normally distributed.

We have the following null and alternative hypothesis

$H_{0}: \mu = 102$ vs $H_{1}: \mu > 102$ (upper-tail alternative)

We will use the test statistic

$Z = \frac{\bar{X}-102}{2/\sqrt{9}}$ and the observed value is

$z_{0} = \frac{100-102}{2/\sqrt{9}} = -3$ .

(a) The rejection region is given by RR = {z | z > 1.6448} where 1.6448 is the 95th quantile of the standard normal distribution. Because the observed value -3 does not belong to RR, we fail to reject the null hypothesis. In other words, there is evidence that the water temperature is acceptable at $\alpha = 0.05$ .

(b) The p-value for this test is given by P(Z > -3) = 0.9987

(c) P(Accepting $H_{0}$ when $\mu = 106$ ) = P(The observed value is not in RR when $\mu = 106$ ) = P( $\frac{\bar{X}-102}{2/\sqrt{9}}$ < 1.6448 when $\mu = 106$ ) = P( $\bar{X}$ < 102 + (1.6448)( $2/\sqrt{9}$ ) when $\mu = 106$ ) = P( $\bar{X}$ < 103.0965) when $\mu = 106$ ) = P( $(\bar{X}-106)/(2/\sqrt{9}) < (103.0965-106)/(2/\sqrt{9})$ )) = P(Z < -4.3552) = 6.647274e-06