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aniked [119]
3 years ago
13

What is the y intercept of f(x)=(1/2)*

Mathematics
1 answer:
9966 [12]3 years ago
5 0

Answer:

B. (0,0)

Step-by-step explanation:

The y-intercept of the equation f(x) = 1/2x lies on the y-axis.

This means we need to set x = 0 and solve for y

1/2(0) = 0

Therefore, (0,0) is the y-intercept of this equation.

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WILL MARK BRANLIEST:)))
zaharov [31]

Answer:

sorry anong grade ka na di kita matutulungan dyan

6 0
2 years ago
Add or subtract. use the least common multiple as the denominator what is 5/8-5/25
Vinvika [58]

Answer:

17/40

Step-by-step explanation:

First let's find the least common denominator. The denominators are 8 and 25 so we need to find the least common multiple of 8 and 25.

8=2*2*2

25=5*5

Since they share no common factors the least common multiple of 8 and 25 is 8*25 which is 200.

Now we convert the fractions:

5/8*25/25=125/200

5/25*8/8=40/200

Then we subtract:

125/200-40/200=85/200

Now we simplify it:

17/40

8 0
2 years ago
Read 2 more answers
9b+5=23 solve equation
yarga [219]

Steps to solve:

9b + 5 = 23

~Subtract 5 to both sides

9b = 18

~Divide 9 to both sides

b = 9

Best of Luck!

5 0
3 years ago
Help with geometry..............................
Ira Lisetskai [31]
(y2-y1)/(x2-x1)
(-7-5)/(4-2)
-12/2= -6
7 0
3 years ago
The manufacturer of an airport baggage scanning machine claims it can handle an average of530 bags per hour.(a-1) At α = .05 in
-Dominant- [34]

Answer:

b. H0: μ < 530. Reject H0 if tcalc > -1.753

Step-by-step explanation:

1) Data given and notation  

\bar X=507 represent the sample mean

s=47 represent the sample standard deviation  

n=16 sample size  

\mu_o =530 represent the value that we want to test  

\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.  

We need to conduct a hypothesis in order to check if the mean is less than 530 (left tailed tes), the system of hypothesis would be:  

Null hypothesis:\mu \geq 530  

Alternative hypothesis:\mu < 530  

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{507-530}{\frac{47}{\sqrt{16}}}=-1.957

Critical value

Since we are conducting a left tailed test we need to first find the degrees of freedom for the statistic given by:

df=n-1=16-1=15

Now we need to look on the t distribution with 15 degrees of freedom that accumulates 0.05 of the area on the left area. And on this case the critical value would be t_{\alpha}=-1/753.

And we can use the following excel code to find it: "=T.INV(0.05,15)"  

So then the correct rejection zone for H0 would be: Reject H0 if t_{calc}

P-value  

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

p_v =P(t_{(15)}  

Conclusion  

If we compare the p value and the significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean is significantly lower than 530 at 5% of significance.  

6 0
3 years ago
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