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2 years ago

8.1g of sugar is needed for every cake made. How much sugar is needed for 6 cakes?

Mathematics

1 answer:

Lerok [7]2 years ago

8 0

Answer:

48.6

Step-by-step explanation:

If you use 8.1g of sugar for 1 cake then 6 cakes will be 48.6g of sugar

Just do 8.1*6 and you will get 48.6

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PLEASE HELP!!!

Arisa [49]

The answer is The root of a number x is another number, which when multiplied by itself a given number of times, equals x. For example the second root of 9 is 3, because 3x3 = 9. The second root is usually called the square root. The third root is susually called the cube root See Root (of a number). because then/

7 0

2 years ago

Read 2 more answers

What is the quotient of StartFraction negative 8 x Superscript 6 Baseline Over 4 x Superscript negative 3 Baseline EndFraction?

scZoUnD [109]

Answer:

Option D.

Step-by-step explanation:

We need to find the quotient of

$\dfrac{-8x^6}{4x^{-3}}$

It can rewritten as

$\dfrac{-8}{4}\times \dfrac{x^6}{x^{-3}}$

$-2\times x^{6-(-3)}$ $[\because \dfrac{a^m}{a^n}=a^{m-n}]$

$-2x^{6+3}$

$-2x^{9}$

The required expression is $-2x^{9}$ .

Therefore, the correct option is D.

8 0

2 years ago

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

Joe hypothesizes that the students of an elite school will score higher than the general population. He records a sample mean eq

Andrew [12]

Answer:

The test to be used is the right tailed test.

Step-by-step explanation:

The type of test joe should do would be a right tailed test. This is because;

A right tailed test which we sometimes call an upper test is where the hypothesis statement contains the greater than (>) symbol. This means that, the inequality points to the right. For example, we want to compare the the life of batteries before and after a manufacturing change.

If we want to know if the battery life of maybe 90 hours would be greater than the original, then our hypothesis statements might be:

Null hypothesis: (H0 = 90).

Alternative hypothesis: (H1) > 90.

In the null hypothesis, there are no changes, but in the alternative hypothesis, the battery life in hours has increased.

So, the most important factor here is that the alternative hypothesis (H1) is what determines if we have a right tailed test, not the null hypothesis.

Thus, the test to be used is the right tailed test.

7 0

2 years ago

Suppose you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by model

Ann [662]

Answer:

(1) The degrees of freedom for unequal variance test is (14, 11).

(2) The decision rule for the 0.01 significance level is;

If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.

If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.

(3) The value of the test statistic is 0.3796.

Step-by-step explanation:

We are given that you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne's attire with those of Calvin Klein.

The following is the amount ($000) earned per month by a sample of 15 Claiborne models;

$3.5, $5.1, $5.2, $3.6, $5.0, $3.4, $5.3, $6.5, $4.8, $6.3, $5.8, $4.5, $6.3, $4.9, $4.2 .

The following is the amount ($000) earned by a sample of 12 Klein models;

$4.1, $2.5, $1.2, $3.5, $5.1, $2.3, $6.1, $1.2, $1.5, $1.3, $1.8, $2.1.

(1) As we know that for the unequal variance test, we use F-test. The degrees of freedom for the F-test is given by;

$\text{F}_(_n__1-1, n_2-1_)$

Here, $n_1$ = sample of 15 Claiborne models

$n_2$ = sample of 12 Klein models

So, the degrees of freedom = ( $n_1-1, n_2-1$ ) = (15 - 1, 12 - 1) = (14, 11)

(2) The decision rule for 0.01 significance level is given by;

If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.

If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.

(3) The test statistics that will be used here is F-test which is given by;

T.S. = $\frac{s_1^{2} }{s_2^{2} } \times \frac{\sigma_2^{2} }{\sigma_1^{2} }$ ~ $\text{F}_(_n__1-1, n_2-1_)$

where, $s_1^{2}$ = sample variance of the Claiborne models data = $\frac{\sum (X_i-\bar X)^{2} }{n_1-1}$ = 1.007

$s_2^{2}$ = sample variance of the Klein models data = $\frac{\sum (X_i-\bar X)^{2} }{n_2-1}$ = 2.653

So, the test statistics = $\frac{1.007}{2.653 } \times 1$ ~ $\text{F}_(_1_4,_1_1_)$

= 0.3796

Hence, the value of the test statistic is 0.3796.

3 0

2 years ago