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3 years ago

I make 2.5kg of popcorn and eat 750g of it while 750g of it while watching a movie . How much popcorn do i have left?

Mathematics

1 answer:

elena-14-01-66 [18.8K]3 years ago

5 0

Answer:

1,75 kg

Explanation:

1 kg = 1000 g

2.5 kg = 2.5 × 1000 g = 2500 g

2500 g - 750 g = 1750 g

1750 g = 1750 ÷ 1000 kg = 1.75 kg

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gogolik [260]

Answer:

Step-by-step explanation:

12x+x+4=17

13x=17-4

13x=13

x=1

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3 years ago

A fabric manufacturer believes that the proportion of orders for raw material arriving late isp= 0.6. If a random sample of 10 o

ryzh [129]

Answer:

a) the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548

- the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504

- the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177

- the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

Step-by-step explanation:

Given the data in the question;

proportion p = 0.6

sample size n = 10

binomial distribution

let x rep number of orders for raw materials arriving late in the sample.

(a) probability of committing a type I error if the true proportion is p = 0.6;

∝ = P( type I error )

= P( reject null hypothesis when p = 0.6 )

= ³∑ $_{x=0$ b( x, n, p )

= ³∑ $_{x=0$ b( x, 10, 0.6 )

= ³∑ $_{x=0$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.6)^x$ $( 1 - 0.6 )^{10-x$

∝ = 0.0548

Therefore, the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548

the probability of committing a type II error for the alternative hypotheses p = 0.3

β = P( type II error )

= P( accept the null hypothesis when p = 0.3 )

= ¹⁰∑ $_{x=4$ b( x, n, p )

= ¹⁰∑ $_{x=4$ b( x, 10, 0.3 )

= ¹⁰∑ $_{x=4$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.3)^x$ $( 1 - 0.3 )^{10-x$

= 1 - ³∑ $_{x=0$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.3)^x$ $( 1 - 0.3 )^{10-x$

= 1 - 0.6496

= 0.3504

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504

the probability of committing a type II error for the alternative hypotheses p = 0.4

β = P( type II error )

= P( accept the null hypothesis when p = 0.4 )

= ¹⁰∑ $_{x=4$ b( x, n, p )

= ¹⁰∑ $_{x=4$ b( x, 10, 0.4 )

= ¹⁰∑ $_{x=4$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.4)^x$ $( 1 - 0.4 )^{10-x$

= 1 - ³∑ $_{x=0$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.4)^x$ $( 1 - 0.4 )^{10-x$

= 1 - 0.3823

= 0.6177

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177

the probability of committing a type II error for the alternative hypotheses p = 0.5

β = P( type II error )

= P( accept the null hypothesis when p = 0.5 )

= ¹⁰∑ $_{x=4$ b( x, n, p )

= ¹⁰∑ $_{x=4$ b( x, 10, 0.5 )

= ¹⁰∑ $_{x=4$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.5)^x$ $( 1 - 0.5 )^{10-x$

= 1 - ³∑ $_{x=0$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$ $(0.5)^x$ $( 1 - 0.5 )^{10-x$

= 1 - 0.1719

= 0.8281

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

3 0

2 years ago

Bill and Annie's ages add up to a total of 30 years. Bill is six years older than twice Annie's age. How old are Bill and Annie?

omeli [17]

Bill is 5 years old and annie is 6 years old so, you multiply 5×6=30

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3 years ago

Solve for x:<br> 3x – 2(6 - x) = 7x + 2(5 + x) – 6

MAVERICK [17]

Answer:

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Step-by-step explanation:

3x-2(6-x)=7x+2(5+x)-6

First, distribute.

3x+ −12+ 2x = 7x + 10 + 2x −6

Combine like terms on each side.

5x − 12= 9x + 4

To solve for x, pick a side to solve. I will solve the right x.

5x - 12 = 9x + 4

-5x -5x

-12 = 4x + 4

Subtract 4 to isolate the variable.

-16 = 4x

Divide 4.

-4 = x

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2 years ago

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maks197457 [2]

Answer:

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Step-by-step explanation:

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2 years ago