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kramer
1 year ago
13

For each 3 mm of coloured fabric Alex uses to make his curtains, he also uses 2 cm of white fabric. Express the amount of white

fabric to coloured fabric as a ratio in its simplest form.
Mathematics
1 answer:
Leokris [45]1 year ago
5 0

Answer: 2cm/.2cm

Step-by-step explanation:

You might be interested in
The percentage of U.S. college freshmen claiming no religious affiliation has risen in recent decades. The bar graph shows the
emmainna [20.7K]

Answer:

0.5%/year

24.2%

Step-by-step explanation:

Estimate the average yearly increase in the percentage of first-year college females claiming no religious affiliation

Percentage of females by year:

1980 = 6.2%

1990 = 10.8%

2000 = 13.6%

2012 = 21.7%

Average yearly increase :

Percentage increase between 1980 - 2012 :

2012% - 1980% = ( 21.7% - 6.2%) = 15.5% increase over [(2012 - 1980)] = 32 years

15.5 % / 32 years = 0.484375% / year = 0.5%/year

b. Estimate the percentage of first-year college females who will claim no religious affiliation in 2030,

Given an average increase of 0.484375% / year

(2030 - 1980) = 50 years

Hence by 2030 ; ( 50 years × 0.484375%/year) = 24.218% will claim no religious affiliation.

=24.2% (nearest tenth)

3 0
2 years ago
Tasha has eight more fish than Oscar, who has twice as many fish as Cecilia. If altogether they have 148 fish, how many fish do
Viefleur [7K]

Answer:

They each have 64

Step-by-step explanation:

let C = number of fish that Cecilia has --- then, O = 2C, where O = number of fish that Oscar has --- and T = O + 8 = 2C + 8, where T = number of fish that Tasha has --- then, C + O + T = C + 2C + (2C + 8) = 148 --- or, 5C + 8 = 148 --- or, 5C = 140 --- or, C = 140/5 = 28 --- and, O = 2C = 2(28) = 56 --- and, T = O + 8 = 56 + 8 = 64

7 0
2 years ago
For every $10 Glen earns, his parents
irga5000 [103]

10-2 = 8

He can spend $8

$8/$10 = 0.80

0.80 x 100 = 80 %

He can spend 80%

8 0
1 year ago
Read 2 more answers
The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100,
Gre4nikov [31]

Answer:

A.the type 1 error probability is \mathbf{\alpha = 0.0244 }

B. β  = 0.0122

C. β  = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

\mathtt{H_o: \mu = 100}

\mathtt{H_1: \mu \neq 100}

A. If the acceptance region is defined as 98.5 <  \overline x >  101.5 , find the type I error probability \alpha .

Assuming the critical region lies within \overline x < 98.5 or \overline x > 101.5, for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is \mu = 100

∴

\mathtt{\alpha = P( type  \ 1  \ error ) = P( reject \  H_o)}

\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5  )}

when  \mu = 100

\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}}  > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }

\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z  > \dfrac{1.5}{\dfrac{2}{3}}) }

\mathtt{\alpha = P ( Z  2.25) }

\mathtt{\alpha = P ( Z

From the standard normal distribution tables

\mathtt{\alpha = 0.0122+( 1-  0.9878) })

\mathtt{\alpha = 0.0122+( 0.0122) })

\mathbf{\alpha = 0.0244 }

Thus, the type 1 error probability is \mathbf{\alpha = 0.0244 }

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis \mathtt{H_o}

Thus;

β = P( type II error) - P( fail to reject \mathtt{H_o} )

∴

\mathtt{\beta = P(98.5 \leq \overline x \leq  101.5)           }

Given that \mu = 103

\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }

\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }

\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }

\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

\mathtt{\beta = P(98.5 \leq \overline x \leq  101.5)           }

Given that \mu = 105

\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }

\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }

\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }

\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

8 0
2 years ago
1. Hugo scored 18 points in a recent basketball game, which was 5 points fewer than Toby scored. Write an equation for this situ
Bingel [31]

Answer: D

Step-by-step explanation:

Is if you do 18-5=13 13 is what Toby scored

6 0
2 years ago
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