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guapka [62]
2 years ago
15

HELLO CAN ANYONE ANSWER THIS PLEASE I WILL MARK YOU THE BRAINLIEST !!!

Mathematics
2 answers:
BartSMP [9]2 years ago
7 0

Answer:

ANGLE CBE=25°

EXPLANATION:

LET ANGLE CBE BE X.

ANGLE CBE = ANGLE ABD

CBE=ABD=X

X+X+155°+155°=360°

2X+310°=360°

X=360-310/2

X=50/2

X=25°

:)

Arte-miy333 [17]2 years ago
5 0
25 degrees is the answer. CBE is supplementary to ABC. That means those two angles add up to 180 degrees.
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2. The product of an even number and an odd number is an even number.
ASHA 777 [7]

Answer:

5

Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
Answer part a and b please and explain thx
Veseljchak [2.6K]
Part A: the area of a rectangle is A= lw, so

A= lw

Height= 6x+3.= 9x

Width= 8

Which makes the expression to this area 9x x 8 ( 9x times 8 )

Part B: you do the same to get the expression here, so the expression would be

6x x 12 ( 6x times 12 )

The answer to this part is yes because both end up being 72x once you get the answer of each.

5 0
2 years ago
I'm a bit confused on how to get x, the example doesn't make sense either.
OLEGan [10]
We know that A=l*w and we also know that these two rectangles have to have the same area. This means we set the two area formulas equal to each other:
(x)(12)=(16)(x-2)
12x=16x-32
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7 0
2 years ago
The mean and standard deviation of a random sample of n measurements are equal to 34.5 and 3.4, respectively.A. Find a 95 % conf
lora16 [44]

Answer:

a. The 95% confidence interval for the mean is (33.52, 35.48).

b. The 95% confidence interval for the mean is (34.02, 34.98).  

c. n=49 ⇒ Width = 1.95

n=196 ⇒ Width = 0.96

Note: it should be a factor of 2 between the widths, but the different degrees of freedom affects the critical value for each interval, as the sample size is different. It the population standard deviation had been used, the factor would have been exactly 2.

d. 5. Quadrupling the sample size while holding the confidence coefficient fixed decreases the width of the confidence interval by a factor of 2.

Step-by-step explanation:

a. We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=34.5.

The sample size is N=49.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.4}{\sqrt{49}}=\dfrac{3.4}{7}=0.486

The degrees of freedom for this sample size are:

df=n-1=49-1=48

The t-value for a 95% confidence interval and 48 degrees of freedom is t=2.011.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.011 \cdot 0.486=0.98

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 34.5-0.98=33.52\\\\UL=M+t \cdot s_M = 34.5+0.98=35.48

The 95% confidence interval for the mean is (33.52, 35.48).

b. We have to calculate a 95% confidence interval for the mean.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.4}{\sqrt{196}}=\dfrac{3.4}{14}=0.243

The degrees of freedom for this sample size are:

df=n-1=196-1=195

The t-value for a 95% confidence interval and 195 degrees of freedom is t=1.972.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=1.972 \cdot 0.243=0.48

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 34.5-0.48=34.02\\\\UL=M+t \cdot s_M = 34.5+0.48=34.98

The 95% confidence interval for the mean is (34.02, 34.98).

c. The width of the intervals is:

n=49\rightarrow UL-LL=33.52-35.48=1.95\\\\n=196\rightarrow UL-LL=34.02-34.98=0.96

d. The width of the intervals is decreased by a factor of √4=2 when the sample size is quadrupled, while the others factors are fixed.

8 0
2 years ago
Is the value of 1,440,000 10 times as great as the value of the 4 in the hundred thousands place?
const2013 [10]
No because 440,000x10= 4,001,000
8 0
3 years ago
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