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8_murik_8 [283]

3 years ago

9

What is 5/6 times 4.

2 answers:

SVETLANKA909090 [29]3 years ago

6 0

The Answer would be 20/6 but that can reduced down to back down to 10/3. To multiply a fraction by a whole number, know that a while number like four is always over 1 so it would 5/6 x 4/1. To multiplay fractions, you multiply the top numbers ( 5 x 4 ) and then your bottom numbers ( 6 x 1 ) which gives you your answer: 20/6 which you might want to reduce if possible which you can.

atroni [7]3 years ago

4 0

Take 5/6 x 4/1 and multiply 5x4 and 6x1 to get 20/6. 6 goes into 20 3 times, with 2/6 leftover. Reduce 2/6 to 1/3 and your final answer is 3 1/3.

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DochEvi [55]

(0.5 liter/bottle) x (6 bottles/pack) = 3 liters/pack

(10 days) x (1.5 liter/day) / (3 liter/pack) = <u>5 packs</u>

8 0

3 years ago

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A math teacher wants to find out the average number of hours his students spend working on their math homework for his class eac

Stolb23 [73]

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4 0

3 years ago

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The mean number of hours of study time per week for a sample of 562 students is 23. If the margin of error for the population me

Vadim26 [7]

Answer:

The 98% confidence interval for the mean number of hours of study time per week for all students is (20.9, 25.1).

Step-by-step explanation:

Confidence interval:

Sample mean plus/minus the margin of error.

In this question:

Mean of 23.

Margin of error 2.1.

Then

23 - 2.1 = 20.9

23 + 2.1 = 25.1

The 98% confidence interval for the mean number of hours of study time per week for all students is (20.9, 25.1).

3 0

2 years ago

WHO CAN solve it Please !

Mariulka [41]

Answer:

a) True

<em> </em> $\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha } )d\alpha =0$ <em></em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given that the definite integration</em>

<em> </em> $\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha } )d\alpha$ <em></em>

<em>we know that the trigonometric formula</em>

<em> sin²∝+cos²∝ = 1</em>

<em> cos²∝ = 1-sin²∝</em>

<u><em>step(ii):-</em></u>

<em>Now the integration</em>

<em> </em> $\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha } )d\alpha = \int\limits^\pi _0 {(\sqrt{cos^{2} \alpha } } \, )d\alpha$ <em></em>

<em> = </em> $\int\limits^\pi _0 {cos\alpha } \, dx$ <em></em>

<em>Now, Integrating </em>

<em> </em> $= ( sin\alpha )_{0} ^{\pi }$ <em></em>

<em> = sin π - sin 0</em>

<em> = 0-0</em>

<em> = 0</em>

<u><em>Final answer:-</em></u>

<em> </em> $\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha } )d\alpha =0$ <em></em>

<em></em>

<em></em>

5 0

3 years ago

I forgot the question

FrozenT [24]

I feel like I’m seeing the same question a lot

6 0

2 years ago