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3 years ago

(15.2*0.25-48.51/14.7)/x=(13/44-2/11-5/66/2.50)1.2/(3.2+0.8(5.5-3.25)

Mathematics

1 answer:

goldfiish [28.3K]3 years ago

6 0

Answer:

x = 41.67

Step-by-step explanation:

The above equation, would be simplified or divided into parts;

Therefore, the given equation becomes;

A/x = B/C

Where;

A = (15.2*0.25-48.51/14.7)

B = (13/44-2/11-5/66/2.50)1.2

C = 3.2+0.8(5.5-3.25)

x = unknown variable.

Part A

(15.2*0.25-48.51/14.7) = (15.2*0.25 - 3.3)

A = (3.8 - 3.3)

A = 0.5

Part B

(13/44-2/11-5/66/2.50)1.2 = (0.3 - 0.18 - 0.030) * 1.2

B = 0.09 * 1.2

B = 0.108

Part C

(3.2+0.8(5.5-3.25)

C = 4*(2.25)

C = 9

Substituting the values into the equation, we have;

0.5/x = 0.108/9

Cross-multiplying, we have;

9 * 0.5 = 0.108x

4.5 = 0.108x

x = 4.5/0.108

x = 41.67

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A store is having a sale on almonds and jelly beans today. The table below shows the amount of each type of food (in pounds) and

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a. The system of equations are:

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12x + 2y = 25

2. Cost for each pound of almond (x) = $1.5

Cost for each pound of jelly beans (y) = $3.5

<h3>How to Solve a System of Equations?</h3>

From the information given above, we can create the system of equations that will be used in solving the problem as shown below.

a. The system of equations that represents the situation is:

3x + 5y = 22 ----> equation 1

12x + 2y = 25 ----> equation 2

b. Multiply equation 1 by 2, and equation 2 by 5:

6x + 10y = 44 ----> equation 3

60x + 10y = 125 ----> equation 4

Subtract equation 4 from equation 3

-54x = -81

-54x/-54 = -81/-54

x = 1.5

Cost for each pound of almond (x) = $1.5

Substitute x = 1.5 into equation 1

3x + 5y = 22 ----> equation 1

3(1.5) + 5y = 22

4.5 + 5y = 22

5y = 22 - 4.5

5y = 17.5

5y/5 = 17.5/5

y = 3.5

Cost for each pound of jelly beans (y) = $3.5

Learn more about the system of equations on:

brainly.com/question/13729904

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1 year ago

A large manufacturer that sells consumer products on-line wishes to publicize its customer satisfaction in an advertisement. Spe

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Answer:

$z=\frac{0.93 -0.9}{\sqrt{\frac{0.9(1-0.9)}{400}}}=2$

$p_v =P(z>2)=0.0228$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who say yes is significantly higher than 0.9 or 90%.

Step-by-step explanation:

Data given and notation

n=400 represent the random sample taken

X=372 represent the number of people who say yes

$\hat p=\frac{372}{400}=0.93$ estimated proportion of people who say yes

$p_o=0.9$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that that more than 90% of its costumers would tell their friend to buy a product from the manufacturer.:

Null hypothesis: $p\leq 0.9$

Alternative hypothesis: $p > 0.9$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.93 -0.9}{\sqrt{\frac{0.9(1-0.9)}{400}}}=2$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level assumed for this case is $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>2)=0.0228$

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