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Vedmedyk [2.9K]
3 years ago
6

2.4 written in simplest form

Mathematics
1 answer:
guajiro [1.7K]3 years ago
7 0
You can either write 2.4 as 12/5 ( an improper fraction) or as 2(2/5) (a mixed number).
You might be interested in
Simplify:<br>3×5-1-(2² +6÷2)​
ivanzaharov [21]

Answer:

7

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Use this information to answer the questions. University personnel are concerned about the sleeping habits of students and the n
Oksanka [162]

Answer:

z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097  

p_v =P(Z>2.097)=0.018  

If we compare the p value obtained and the significance level given \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  students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

Step-by-step explanation:

1) Data given and notation

n=377 represent the random sample taken

X=209 represent the students reported experiencing excessive daytime sleepiness (EDS)

\hat p=\frac{209}{377}=0.554 estimated proportion of students reported experiencing excessive daytime sleepiness (EDS)

p_o=0.5 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 the true proportion is higher than 0.5:  

Null hypothesis:p\leq 0.5  

Alternative hypothesis:p > 0.5  

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{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097  

4) 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 provided \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>2.097)=0.018  

If we compare the p value obtained and the significance level given \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  students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

6 0
3 years ago
In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A
vredina [299]

Answer:

c) P(270≤x≤280)=0.572

d) P(x=280)=0.091

Step-by-step explanation:

The population of bearings have a proportion p=0.90 of satisfactory thickness.

The shipments will be treated as random samples, of size n=500, taken out of the population of bearings.

As the sample size is big, we will model the amount of satisfactory bearings per shipment as a normally distributed variable (if the sample was small, a binomial distirbution would be more precise and appropiate).

The mean of this distribution will be:

\mu_s=np=500*0.90=450

The standard deviation will be:

\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7

We can calculate the probability that a shipment is acceptable (at least 440 bearings meet the specification) calculating the z-score for X=440 and then the probability of this z-score:

z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932

Now, we have to create a new sampling distribution for the shipments. The size is n=300 and p=0.932.

The mean of this sampling distribution is:

\mu=np=300*0.932=279.6

The standard deviation will be:

\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36

c) The probability that between 270 and 280 out of 300 shipments are acceptable can be calculated with the z-score and using the continuity factor, as this is modeled as a continuos variable:

P(270\leq x\leq280)=P(269.5

d) The probability that 280 out of 300 shipments are acceptable can be calculated using again the continuity factor correction:

P(X=280)=P(279.5

8 0
3 years ago
PLEASE PLEASE HELP
marta [7]

Answer:

Approximately 3 grams left.

Step-by-step explanation:

We will utilize the standard form of an exponential function, given by:

f(t)=a(r)^t

In the case of half-life, our rate <em>r</em> will be 1/2. This is because 1/2 or 50% will be left after <em>t </em>half-lives.

Our initial amount <em>a </em>is 185 grams.

So, by substitution, we have:

\displaystyle f(t)=185\big(\frac{1}{2}\big)^t

Where <em>f(t)</em> denotes the amount of grams left after <em>t</em> half-lives.

We want to find the amount left after 6 half-lives. Therefore, <em>t </em>= 6. Then using our function, we acquire:

\displaystyle f(6)=185\big(\frac{1}{2}\big)^6

Evaluate:

\displaystyle f(6)=185\big(\frac{1}{64}\big)\approx2.89\approx 3

So, after six half-lives, there will be approximately 3 grams left.

3 0
2 years ago
Snowy's Snow Cones has a special bubble gum snow cone on sale. The cone is a regular snow cone that has a spherical plece of bub
Aloiza [94]

Draw a diagram to illustrate the problem as shown below.

Calculate the volume of the empty cone.

V₁ = (1/3)π*(6 in)²*(10 in) = 120π in³

Calculate the volume of the sphere.

V₂ = (4/3)π*(1.5 in)³ = 4.5π in³

The volume that can be filled with flavored ice is

V = V₁ - V₂ = 115.5π in³ = 362.85 in³

Answer:

The volume is 115.5π in³ or 362.9 in³ (nearest tenth)

4 0
2 years ago
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