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Sonja [21]
2 years ago
10

Q7: Express each of the following: a) 4.5m2 in cm2

Mathematics
1 answer:
Readme [11.4K]2 years ago
8 0

1m2 = 10000cm2

4.5m2 = 25000cm2

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A line passes through (2,8) and (4,12) which equation best represents the line
Dmitry [639]

Answer:

y=2x+4

Step-by-step explanation:\solving for the slope you get (12-8)/(4-2)=4/2=2

solving for the y intercept you subtract 4 from 8 because 2*2 is 4 and (2,8) is 2 x values away from the y intercept so 8-4=4 so the y intercept is 4

so y=2x+4

4 0
3 years ago
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Is the relation a function why or why not???
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Answer:

No, it is not a function. The x repeats.

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2 years ago
A cigarette industry spokesperson remarks that current levels of tar are no more than 5 milligrams per cigarette. A reporter doe
d1i1m1o1n [39]

Answer:

t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516  

df=n-1=15-1=14  

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

p_v =P(t_{14}>1.516)=0.076  

If we use a significance level of 0.05 we see that p_v > \alpha and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

Step-by-step explanation:

Data given and notation  

\bar X=5.63 represent the sample mean  

s=1.61 represent the standard deviation for the sample  

n=15 sample size  

\mu_o =15/tex] represent the value that we want to test  
[tex]\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the average is more than 5.63, the system of hypothesis would be:  

Null hypothesis:\mu \leq 5.63  

Alternative hypothesis:\mu > 5.63  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

t=\frac{5.63-5}{\frac{1.61}{\sqrt{15}}}=1.516  

Now we need to find the degrees of freedom for the t distribution given by:

df=n-1=15-1=14  

P value

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

p_v =P(t_{14}>1.516)=0.076  

If we use a significance level of 0.05 we see that p_v > \alpha and then we can conclude that we don't have evidence in order to conclude that the mean is higher than 5.63, so then the claim makes sense.

3 0
3 years ago
The function g(x) is shown in the graph at right. The function g(x) is a transformation of the parent function f(x)=x^2. Based o
Ivahew [28]

Answer: choice B

Step-by-step explanation:

6 0
2 years ago
Consider the continuous random variable x, which has a uniform distribution over the interval from 110 to 150. The probability t
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Answer:

The probability that x will take on a value between 120 and 125 is 0.14145

Step-by-step explanation:

For uniform distribution between a & b

Mean, xbar = (a + b)/2

Standard deviation, σ = √((b-a)²/12)

For 110 and 150,

Mean, xbar = (150 + 110)/2 = 130

Standard deviation, σ = √((150-110)²/12 = 11.55

To find the probability that x will take on a value between 120 and 125

We need to standardize 120 & 125

z = (x - xbar)/σ = (120 - 130)/11.55 = - 0.87

z = (x - xbar)/σ = (125 - 130)/11.55 = - 0.43

P(120 < x < 125) = P(-0.87 < x < -0.43)

We'll use data from the normal probability table for these probabilities

P(120 < x < 125) = P(-0.87 < x < -0.43) = P(z ≤ -0.43) - P(z ≤ -0.86) = 0.33360 - 0.19215 = 0.14145

Hope this Helps!!!

3 0
3 years ago
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