All categories

2 years ago

Change mixed numbers into improper fraction. 8 3/6, 4 3/8, 2 2/4, 9 4/5, 7 2/3

Mathematics

1 answer:

andrew-mc [135]2 years ago

8 0

Answer: 8 3/6= 17/2

4 3/8=35/8

2 2/4=5/2

9 4/5 =49/5

7 2/3= 23/3

Step-by-step explanation:

To convert a mixed number to a improper fraction you multiply the whole number part by the fractions denominator then you add that number to the numerator then you write the number on top of the denominator

You might be interested in

Which model represents the factors of 4x2 – 9? An algebra tile configuration. 5 tiles are in the Factor 1 spot: 2 +x , 3 negativ

Arturiano [62]

Answer:

Step-by-step explanation:

4x² - 9

(2x)² - 3²

(2x + 3)(2x - 3)

Factor 1 spot:

2 tiles of x

3 tiles of positive 1

Factor 2 spot:

2 tiles of x

3 tiles of negative 1

5 0

2 years ago

Read 2 more answers

A clock is constructed using a regular polygon with 60 sides. The polygon rotates each minute, making one full revolution each h

Maksim231197 [3]

Answer:

D. 42

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Mr. Davies spent $240 on lunch for his class. Sandwiches x cost $6 and drinks y cost $2. This can be represented by the equation

gtnhenbr [62]

Answer: the x intercept represents the maximum number of sandwiches that can be purchased bc no drinks would be purchased. The y intercept represents how many drinks could be purchased if he didn’t buy any sandwiches

Step-by-step explanation:

7 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

The point (Negative StartFraction StartRoot 2 EndRoot Over 2 EndFraction, StartFraction StartRoot 2 EndRoot Over 2 EndFraction)

Ierofanga [76]

Answer:

$\cot(x) = - 1 \: and \: \cos(x) = - \frac{ \sqrt{2} }{2}$