S + 2t + 39= t + 3s + 30 Find S

Which of the following values do not satisfy the inequality -2x-6<_1

Feliz [49]

Set of numbers
R-[-7/2,infinity)
symbols have there usual mean

3 0

3 years ago

What is this fraction in decimal form? 3/8 Enter your answer in the box.

xxTIMURxx [149]

it is 0.375

happy learning

4 0

2 years ago

Complete the equation involving x. Order terms like the physical situation, and don't simplify. A 6 foot wide painting should be

34kurt

Answer:

The space on each side of the painting = 2 feet

Step-by-step explanation:

Attached this answer is a diagram to give you visuals on how the value of x is calculated

Given:

width of wall = 10 feet

width of painting = 6 feet

Since the painting is centred on a wall, there is meant to be an equal space on each side of the painting, from the diagram, the following eaquation can be gotten

x + 6 + x = 10

2x + 6 = 10

2x = 10 - 6

2x = 4

x = 4 ÷2

x = 2 feet

Therefore, the space on each side of the painting = 2 feet

8 0

4 years ago

Find the total volume

Elodia [21]

Answer:

My calculation might be wrong, so take A.

Step-by-step explanation:

Volume= Volume of hemisphere+volume of cone

Volume= $\frac{2}{3}$ ×π×r³ + πr² $\frac{h}{3}$

Volume= $\frac{2}{3}$ × $\frac{22}{7}$ ×4×4×4 + $\frac{22}{7}$ ×4×4× $\frac{14}{3}$

Volume= 134.095 + 410.6

Volume= 544.695

4 0

3 years ago

It has been reported that the probability that an individual will develop schizophrenia over their lifetime is 0.004. In a rando

Alinara [238K]

Answer:

Null hypothesis: $p=0.004$

Alternative hypothesis: $p \neq 0.004$

$z=\frac{0.00567 -0.004}{\sqrt{\frac{0.004(1-0.004)}{3000}}}=1.449$

$p_v =2*P(z>1.149)=0.2506$

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion if individuals that developed schizophrenia its not significantly different from 0.004.

Step-by-step explanation:

1) Data given and notation

n=3000 represent the random sample taken

X=17 represent the individuals developed schizophrenia in the sample

$\hat p=\frac{17}{3000}=0.00567$ estimated proportion of individuals developed schizophrenia in the sample

$p_o=0.004$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportion of people who will develop schizophrenia is different from 0.004:

Null hypothesis: $p=0.004$

Alternative hypothesis: $p \neq 0.004$

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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.00567 -0.004}{\sqrt{\frac{0.004(1-0.004)}{3000}}}=1.449$

4) P value

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 provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(z>1.149)=0.2506$

5) Decision

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion if individuals that developed schizophrenia its not significantly different from 0.004.

4 0

4 years ago