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3 years ago

Express the fraction 3/4, 7/16 and 5/8 with the LCD

1 answer:

3 0

12/16, 7/16, 10/16 (not simplified, just with LCD);The LCD of the 3 fractions is 16 (from denominators) so all we have to do is multiply both the numerator by the same number that multiplies with the denominators to get 16. We leave 7/16 alone since it already has denominator of 16.

3/4 * 4 = 12/16

5/8 * 2 = 10/16

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It costs $26 to fertilize, water, mow, and maintain each square yard of a full size FIFA field (with maximum dimensions) before

ElenaW [278]

Answer:

It will cost $256542 to prepare the field for next weeks match.

Step-by-step explanation:

The maximum dimensions of a field according to FIFA rules is 110 m by 75 m.

So, the maximum area of a field is (110 × 75) = 8250 square meters.

Now, 1 square meters is equivalent to 1.196 square yards.

So, the maximum area of a field is (8350 × 1.196) = 9867 square yards.

Now, given that it costs to make ready each square yard of a full-size FIFA field, $26 before each game.

Therefore, it will cost $(26 × 9867) = $256542 to prepare the field for next week's match. (Answer)

4 0

3 years ago

A confidence interval (CI) is desired for the true average stray-load loss u (watts) for a certain type of induction motor when

Genrish500 [490]

Answer:

A) CI = (57.12 , 59.48)

B) CI = (57.71 , 58.89)

C) CI = (57.53 , 59.07)

D) n = 239.63

Step-by-step explanation:

given data:

mean, $\bar X = 58.3$

standard deviation, σ = 3

sample size, n = 25Given CI level is 95%, hence α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025,

Zc = Z(α/2) = 1.96

$ME = Zc * σ \sqrt{n}$

$ME = 1.96 * 3 \sqrt{25}$

ME = 1.18

$CI = (\bar X - Zc * s\sqrt{n} , \barX + Zc * s\sqrt{n})$

$CI = (58.3 - 1.96 * 3\sqrt{25} , 58.3 + 1.96 * 3\sqrt{25})$

CI = (57.12 , 59.48)

Given data:

mean, $\bar X = 58.3$

standard deviation, σ = 3

sample size, n = 100

Given CI level is 95%, hence α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

$ME = zc * σ \sqrt{n}$

$ME = 1.96 * 3\sqrt{100}$

ME = 0.59

$CI = (\bar X - Zc * s\sqrt{n} , \barX + Zc * s\sqrt{n})$

$CI = (58.3 - 1.96 * 3\sqrt{100} , 58.3 + 1.96 * 3\sqrt{100})$

CI = (57.71 , 58.89)

sample mean, $\bar X = 58.3$

sample standard deviation, σ = 3

sample size, n = 100

Given CI level is 99%, hence α = 1 - 0.99 = 0.01

α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

$ME = Zc * σ \sqrt{n}$

$ME = 2.58 * 3\sqrt{100}$

ME = 0.77

$CI = (\bar X - Zc * s\sqrt{n} , \barX + Zc * s\sqrt{n})$

$CI = (58.3 - 2.58 * 3\sqrt{100} , 58.3 + 2.58 * 3/\sqrt{100}$

CI = (57.53 , 59.07)

Given data:

Significance Level, α = 0.01,

Margin or Error, E = 0.5,

σ = 3

The critical value for α = 0.01 is 2.58.

for calculating population mean we used

$n \geq (zc *σ/E)^2$

$n = (2.58 * 3/0.5)^2$

n = 239.63

7 0

3 years ago

The giraffes roam 612 acres of grasslands and the tigers roam 346 few acres. How many acres do they roam combined?

jeyben [28]

Answer:

They roam 958 acres combined.

Step-by-step explanation:

612 + 346

= 958

4 0

2 years ago

What -7 x4=36? amogus amogus amogus amogus

77julia77 [94]

Based on the given problem of -7x⁴ = 36, the solution to the equation is x = -1.5.

<h3>What is the solution to this problem?</h3>

The equation to find x is given as:

-7x⁴ = 36

Solving it gives:

-7x⁴ = 36

x⁴ = 36/-7

x⁴ = -36/7

x = ⁴√(-36/7)

x = -1.5

In conclusion, x is -1.5.

Find out more questions on solving for x at brainly.com/question/2910769

#SPJ1

3 0

1 year ago

Tammy conducted a survey to find the favorite subject of the students at her school. She asked 25 students from her math class w

almond37 [142]

Tammy's sample may not be considered valid because, on the first hand, it is said that she only asked students from her " Math Class".
If she wants to have a survey to find out the favorite subject of the students at her school, she must conduct a survey involving all the students in her school, not just in her class. What she did is just subjective. She should use a tally listing the different subjects and compare the number of students per subject. This way, she can have an objective representation of the least liked subjects and the most liked subjects of the students on her school.
Illustrating her survey through statistics may be more reliable and valid because it shows frequencies in which she can calculate easily and accurately the percentage of the number of students per subject, in a more objective manner.

7 0

3 years ago