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Nady [450]
3 years ago
8

Given circle o, what is the length of arc AB if m AOB =40° ? Radius is 9

Mathematics
1 answer:
lorasvet [3.4K]3 years ago
7 0

Answer:

Arc length = 6.28 units

Step-by-step explanation:

Arc \ length = 2 \pi r \frac{ \theta}{360} = 2 \times \pi \times 9 \times \frac{40}{360} = 2\pi = 6.28

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A random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test. The
disa [49]

Answer:

The 90 % confidence limits are (-2.09, 8.09).

Since the calculated values do not lie in the critical region we accept our null hypothesis.

Step-by-step explanation:

The null and alternative hypothesis are given by

H0: σ₁²= σ₂² against Ha: σ₁² ≠ σ₂²

Confidence interval for the population mean difference is given by

(x`1- x`2) ± t √S²(1/n1 + 1/n2)

Where S ²= (n1-1)S₁² + S²₂(n2-1)/n1+n2-2

Critical value of t with n1+n2-2= 50+ 35-2= 83 will be -1.633

Now calculating

S ²=34* (12.8)²+ (14.6)²*49/83= 192.96

Now putting the values in the t- test

(75.1 -72.1) ± 1.633 √ 192.96(1/35 +1/50)

=3 ±  5.09

=-2.09, 8.09  is the 90 % confidence interval for the difference

The 90 % confidence limits are (-2.09, 8.09).

Since the calculated values do not lie in the critical region we accept our null hypothesis.

5 0
3 years ago
a jar containing 100 pennies is dumped onto a table, and all those coming up heads are removed. The remaining pennies are placed
lakkis [162]

Sorry, but there is no table to the right. Please put the table.

8 0
3 years ago
-19+3x-11+2x=2<br> What the answer
Vladimir [108]

Answer:

Step-by-step explanation:

x = 32/5

Decimal Form:

x = 6.4

Mixed Number Form:

x = 6 2\5

8 0
3 years ago
Read 2 more answers
Which of the following equations is an example of inverse variation between variables x and y
Firdavs [7]

Answer:

option A y=9/x

Step-by-step explanation:

It is said that two variables x and y vary inversely if the increase of one of the variables causes the other to decrease.

3 0
3 years ago
A group of students wanted to investigate the claim that the average number of text messages sent yesterday by
Likurg_2 [28]
First we need to write the null and alternate hypothesis for this case.

Let x be the average number of text message sent. Then

Null hypothesis: x = 100
Alternate hypothesis: x > 100

The p value is 0.0853

If p value > significance level, then the null hypothesis is not rejected. If p value < significance level, then the null hypothesis is rejected.

If significance level is 10%(0.10), the p value will be less than 0.10 and we reject the null hypothesis and CAN conclude that:
The mean number of text messages sent yesterday was greater than 100.

If significance level is 5%(0.05), the p value will be greater than 0.05 and we cannot reject the null hypothesis and CANNOT conclude that:
The mean number of text messages sent yesterday was greater than 100.
7 0
3 years ago
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