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polet [3.4K]
2 years ago
14

Look at the pic below an help pls

Mathematics
1 answer:
GarryVolchara [31]2 years ago
4 0

Answer:

y=2/3(x-3)^2+1

Step-by-step explanation:

The parent function of a parabola is y=x^2:

The transformation of this graph shows the parabola is shifted to the right 3 and up 1 so we apply that to our equation y=(x-3)^2+1

to solve the stretch factor:

y=a(x-3)^2+1

we can see when x=0 y=7 so we use this set

7=a(0-3)^2+1

7=a(-3)^2+1

-1            -1

6=9a

divide by 9

a=2/3

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The GCF of 10 and 18 is 6.<br> TRUE<br> FALSE
FinnZ [79.3K]

Answer:

False

Step-by-step explanation:

Because the gcf of 10 and 18 is 2

7 0
3 years ago
Read 2 more answers
According to a 2010 study conducted by the Toronto-based social media analytics firm Sysomos, 71% of all tweets get no reaction.
klemol [59]

Answer:

a) the expected number of tweets with no reaction is 71 tweets

b) the variance is 20.59 tweets² ( ≈ 21 tweets )and the standard deviation is

4.537 tweets (≈ 5 tweets)

Step-by-step explanation:

we can use the binomial probability distribution

P(x, N, p ) = N!/[(N-x)!x!] * p^x * (1-p)^(N-x)

where

x = number of successful events

N = population total

p = probability for success for every individual and independent event

P = probability for x successful events

in our case p = 71% = 71/100 (probability of a tweet without reaction) , N = 100 tweets

a) the expected number of tweets

E(x) = N* p = 100 tweets * 71/100 = 71 tweets

b) the variance is

V(x) = N * p * (1-p)

V = 100 * 71/100 * (1-71/100) = 71* 29/100 = 20.59 tweets² ( units of variance are [N²] )

the standard deviation is

s = √V = √20.59 tweets² = 4.537 tweets

6 0
3 years ago
Frank has a batting average of .255. Write this decimal as a fraction in simplest form.
Artist 52 [7]
9/40 because when you put it in fraction form its 225/1000 then you simplify!

6 0
3 years ago
Read 2 more answers
What type of insurance can be waived (refused) if a driver feels that his/her car is not worth insuring? A. collision B. compreh
Mars2501 [29]

Answer:

i believe its a

Step-by-step explanation:

5 0
2 years ago
A researcher believes that the mean weight of competitive runners is about 140 pounds. A sample of 24 elite distance runners has
ExtremeBDS [4]

Answer:

t=\frac{136-140}{\frac{11}{\sqrt{24}}}=-1.781    

p_v =P(t_{(23)}  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=136 represent the sample mean

s=11 represent the sample standard deviation

n=24 sample size  

\mu_o =140 represent the value that we want to test

\alpha=0.01 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.  

We need to conduct a hypothesis in order to check if the mean is less than 140, the system of hypothesis would be:  

Null hypothesis:\mu \geq 140  

Alternative hypothesis:\mu < 140  

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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".  

Calculate the statistic

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

t=\frac{136-140}{\frac{11}{\sqrt{24}}}=-1.781    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=24-1=23  

Since is a one left tailed test the p value would be:  

p_v =P(t_{(23)}  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.  

4 0
3 years ago
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