True or false: f(x) is a function A. True B. False

What's the greats common factor of 28 42x

LuckyWell [14K]

I think the answer is 7

5 0

3 years ago

Read 2 more answers

4.1(-7.93-4.39a)-4.6a

ki77a [65]

Answer:

−22.599a−32.513

Step-by-step explanation:

Distribute:

= (4.1) (−7.93) + (4.1) (− 4.39 a) + − 4.6 a

= − 32.513 + − 17.999 a + − 4.6 a

Combine Like Terms:

= −32.513 + − 17.999 a + − 4.6 a

= (− 17.999 a + −4.6 a) + (− 32.513)

= − 22.599 a + (− 32.513) = − 22.599 a − 32.513

3 0

3 years ago

Read 2 more answers

Evaluate the function at x = -2.

Lera25 [3.4K]

Answer:

B. y=0

Step-by-step explanation:

7 0

3 years ago

In a recent year, a poll asked 2362 random adult citizens of a large country how they rated economic conditions. In the poll,

Harman [31]

Answer:

a) The 99% confidence interval is given by (0.198;0.242).

b) Based on the p value obtained and using the significance level assumed $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we fail to reject the null hypothesis, and we can said that at 1% of significance the proportion of people who are rated with Excellent/Good economy conditions not differs from 0.24. The interval also confirms the conclusion since 0.24 it's inside of the interval calculated.

c) $\alpha=0.01$

Step-by-step explanation:

Data given and notation

n=2362 represent the random sample taken

X represent the people who says that they would watch one of the television shows.

$\hat p=\frac{X}{n}=0.22$ estimated proportion of people rated as Excellent/Good economic conditions.

$p_o=0.24$ 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 24% of people are rated with good economic conditions:

Null hypothesis: $p=0.24$

Alternative hypothesis: $p \neq 0.24$

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 $p_o$ .

Part a: Test the hypothesis

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

np = 2362x0.22=519.64>10 and n(1-p)=2364*(1-0.22)=1843.92>10

Condition satisfied.

Calculate the statistic

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

$z=\frac{0.22 -0.24}{\sqrt{\frac{0.24(1-0.24)}{2362}}}=-2.28$

The confidence interval would be given by:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

The critical value using $\alpha=0.01$ and $\alpha/2 =0.005$ would be $z_{\alpha/2}=2.58$ . Replacing the values given we have:

$0.22 - (2.58)\sqrt{\frac{0.22(1-0.22)}{2362}}=0.198$

$0.22 + (2.58)\sqrt{\frac{0.22(1-0.22)}{2362}}=0.242$

So the 99% confidence interval is given by (0.198;0.242).

Part b

Statistical decision 

P value method or p value approach . "This method consists on 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 provided is $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z$

So based on the p value obtained and using the significance level assumed $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we fail to reject the null hypothesis, and we can said that at 1% of significance the proportion of people who are rated with Excellent/Good economy conditions not differs from 0.24. The interval also confirms the conclusion since 0.24 it's inside of the interval calculated.

Part c

The confidence level assumed was 99%, so then the signficance is given by $\alpha=1-confidence=1-0.99=0.01$

6 0

3 years ago

A chemist whishes to prepare 100 liters of 45% purity of sulphuric acid .He has two kinds of acid solutions in stock ,one is 55%

tester [92]

Answer:

the chemist should use 60 liters of 55% solution and 40 litres of 30% solution in order to prepare 100 liters of 45% purity of sulphuric acid.

Step-by-step explanation:

From the given information,

Let x be the litres of 55% pure solution

Let y be the litres of 30% pure solution

Also;

Given that our total volume of solution is 100 litres

x+y =100 ---- (1)

The total solution of pure by related by the sum of the individual pure concentrations to make up the concentration of final solution.

(0.55)(x)+(0.30)(y) = 0.45(100) ---- (2)

From equation (1)

Let ; y = 100 - x

Replacing the value for y = 100 - x into equation (2)

(0.55)(x)+(0.30)(100-x) = 0.45(100)

0.55x + 30 - 0.30x = 45

0.55x - 0.30x = 45 - 30

0.25x = 15

x = 15/0.25

x = 60 liters of 55% solution

From ; y = 100 - x

y = 100 - 60

y = 40 litres of 30% solution.

Therefore, the chemist should use 60 liters of 55% solution and 40 litres of 30% solution in order to prepare 100 liters of 45% purity of sulphuric acid.

7 0

4 years ago