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igor_vitrenko [27]
3 years ago
14

Ruben bought 2 large popcorn buckets and 4 small drinks for $20. Lucia bought 2

Mathematics
1 answer:
Korolek [52]3 years ago
7 0

Answer:

4$ for large 3$ for a small

Step-by-step explanation:

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What is the answer to this----->7/8-5/6
dlinn [17]
<span>7/8-5/6
Multiply until both fractions have the same common denominator.
21/24 - 20/24
Subtract the numerators but NOT the denominators.
Final Answer: 1/24</span>
3 0
4 years ago
Read 2 more answers
A tub is filled with 18000 cm3 of water. It loses one-tenth of its water every 12 hours. What volume of water would be remaining
Lina20 [59]

Answer:

0 cm³

Step-by-step explanation:

the lost of water =

(5×24)/12 x 1/10 × 18000

= 10/10 × 18000 = 18,000 cm³

so, the remaining water in the tub=

18,000 - 18,000 = 0 cm³

(the tub is empty)

6 0
3 years ago
I have a 3 foot board that needs to be equally decided into 1/2 how many extra peices will I have
anygoal [31]

Answer:

No extra pieces.

Step-by-step explanation:

Given:

Total number of foot board = 3

Equally divided into =  1/2

Question asked:

How many extra pieces will I have = ?

Solution:

By dividing 3 by 1/2 we can find the extra pieces.

                                                                 = 3\div\frac{1}{2}

                                                                 = 3\times\frac{2}{1}

                                                                 = 6

Since, the remainder is zero, thus there is no any extra pieces I have.

7 0
3 years ago
Provide a counterexample:
Jobisdone [24]

Answer:

a=2 \quad \text{and} \quad b=-2

Step-by-step explanation:

Take a=2 \quad \text{and} \quad b=-2, note that

2=(-1)\cdot(-2)

hence b divides a. On the other hand, we have that

-2=(-1)\cdot2

which tells us that a divides b. Moreover, a=2 \neq -2=b.

5 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:

a)

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)

b)

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)

c)

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)

D)

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
4 years ago
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