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padilas [110]
3 years ago
11

What is happening to this graph when the x values are between -1 and 1

Mathematics
2 answers:
arsen [322]3 years ago
5 0
It is decreasing because the line goes downward from left to right in the area between -1 and 1.

Hope this helps!
Amiraneli [1.4K]3 years ago
3 0
It is decreasing, since the y-value is going down.
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HELP PLEASE ONE QUESTION I DONT GET IT 6th GRADE MATH
grigory [225]

Answer:

-2.65

Step-by-step explanation:

-1.2 - 1.45 = -2.65

you could just simplify fraction into decimal and use calculator

8 0
3 years ago
Read 2 more answers
The area of a circle is 132.7 square cm. Find the diameter.
hram777 [196]

13 cm (approximately)

Step-by-step explanation:

Let the radius of the circle be 'r'.

Area of the circle = πr²

=> 132.7 = (22/7) r²

=> 6.5 ≈ r

Therefore, diameter = 2radius = 2(6.5) = 13 cm

6 0
2 years ago
3-2y+(-8y)+8.5<br><br> Simplify your answer. Use interfere or decimals for any number expressions
Nataly_w [17]
Im pretty sure it’s 11.5 - 16y
8 0
3 years ago
Granola Crunch cereal is packaged in 1 pound boxes. Susan Torres, a quality control analyst who works for the manufacturer of th
sineoko [7]

Answer:

z=\frac{1.02-1}{\frac{0.04}{\sqrt{40}}}=3.162    

p_v =2*P(z>3.162)=0.0016    

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to to reject the null hypothesis, so we can conclude that the true mean is different from 1 pound at 5% of signficance.

Step-by-step explanation:

Data given and notation    

\bar X=1.02 represent the sample mean

\sigma=0.04 represent the population standard deviation    

n=40 sample size    

\mu_o =1 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.    

z 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 true mean is equal to 1 pound or no :    

Null hypothesis:\mu =1    

Alternative hypothesis:\mu \neq 1    

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:    

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)    

z-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:    

z=\frac{1.02-1}{\frac{0.04}{\sqrt{40}}}=3.162    

P-value    

Since is a two-sided test the p value would be:    

p_v =2*P(z>3.162)=0.0016    

Conclusion    

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to to reject the null hypothesis, so we can conclude that the true mean is different from 1 pound at 5% of signficance.    

5 0
3 years ago
two telephone calls come into a switchboard at times that are uniformly distributed in a fixed one-hour period. assume that the
AleksandrR [38]

We wil assume a variable x to be the total number of calls received by the switchboard.

The question also says to assume that the calls were made independently.

Given:

Calls are independent.

Calls are uniformly distributed over a 1 hour period.

Ans (a). The calls are distributed uniformly over 1 hour, hence: (0, 1).

So we have,

f1(x1) = 1

f2(x2) = 1

X1 and X2 are considered to be independent of each other. Hence,

f(x1,x2) = f1 (x1) f2 (x2)

f(x1,x2) = 1 (1)

f(x1,x2) = 1

Thus,

P(X1 <= 0.5; X2 <= 0.5) = ∫0.50 ∫0.50 f(x1,x2) dx2 dx1

= ∫^0.5 0 ∫^0.5 0 (1) dx2 dx1

= ∫^0.5 0 (x2^0.5 0) dx1

= ∫^0.5 0 (0.5 - 0) dx1

= 0.5 ∫^0.5 0 dx1

= 0.5 (x1^0.5 0)

= 0.5 (0.5 - 0)

= 0.25

Therefore, the probability that the calls were received within the first 30 minutes or first half hour is 0.25.

Ans (b). Steps 1 and 2 are the same as the above answer.

Probability = [∫^11/12 0 ∫^x1 + 1/12 x1 1dx2 dx1] + [∫^1 1/12 ∫^x1 x1-1/12 1dx2 dx1

= [∫^11/12 0 (x2 ^x1+1/12 x1 dx1] + [∫^1 1/12 (x2 ^x1 x1-1/12 dx1)]

= [∫^11/12 0 (x1 + 1/12 -x1) dx1] + [∫^1 1/12 (x1 - x1 + 1/12) dx1]

= [(1 + x1/12 - 1) ^11/12 0] + [( 1 - 1 + x1/12) ^1 1/12]

= 11/144 + 11/144

= 0.1528

Therefore, the probability that the calls were received within five minutes of each other is 0.15.

Find more from: brainly.com/question/18125359?referrer=searchResults

#SPJ4

7 0
1 year ago
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