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aivan3 [116]
3 years ago
7

Find the Difference

Mathematics
1 answer:
Alex777 [14]3 years ago
4 0
The difference of these is 30,10,-10
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The area of a circle is 30.85cm2. Find the length of the radius rounded to 2 DP
Alex17521 [72]

Answer: 3.13

Step-by-step explanation:

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2 years ago
Aidan rollerblades 540 ft per min. at this rate how far in yards will he rollerblade in 5 min( 1 yd =3 ft)​
Pani-rosa [81]

Answer:

900 yards

Step-by-step explanation:

540/3=180

180x5=900

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3 years ago
Whats the average between 95 and 128
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Below!

Step-by-step explanation:

111.5

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2 years ago
Use the sample information 11formula13.mml = 37, σ = 5, n = 15 to calculate the following confidence intervals for μ assuming th
bija089 [108]

Answer & Step-by-step explanation:

The confidence interval formula is:

I (1-alpha) (μ)= mean+- [(Z(alpha/2))* σ/sqrt(n)]

alpha= is the proposition of the distribution tails that are outside the confidence interval. In this case, 10% because 100-90%

σ= standard deviation. In this case 5

mean= 37

n= number of observations. In this case, 15

(a)

Z(alpha/2)= is the critical value of the standardized normal distribution. The critical valu for z(5%) is 1.645

Then, the confidence interval (90%):

I 90%(μ)= 37+- [1.645*(5/sqrt(15))]

I 90%(μ)= 37+- [2.1236]

I 90%(μ)= [37-2.1236;37+2.1236]

I 90%(μ)= [34.8764;39.1236]

(b)

Z(alpha/2)= Z(2.5%)= 1.96

Then, the confidence interval (90%):

I 95%(μ)= 37+- [1.96*(5/sqrt(15)) ]

I 95%(μ)= 37+- [2.5303]

I 95%(μ)= [37-2.5303;37+2.5303]

I 95%(μ)= [34.4697;39.5203]

(c)

Z(alpha/2)= Z(0.5%)= 2.5758

Then, the confidence interval (90%):

I 99%(μ)= 37+- [2.5758*(5/sqrt(15))

I 99%(μ)= 37+- [3.3253]

I 99%(μ)= [37-3.3253;37+3.3253]

I 99%(μ)= [33.6747;39.3253]

(d)

C. The interval gets wider as the confidence level increases.

8 0
3 years ago
If the function f(x) = (2x - 3) is transformed to g(x) = (-2x - 3), which type of transformation occurred?
lidiya [134]
C




Ur final answer is C
4 0
3 years ago
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