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serious [3.7K]
3 years ago
12

Find b. Round to the nearest tenth:

Mathematics
2 answers:
igor_vitrenko [27]3 years ago
6 0

Answer:

6.7 cm

Step-by-step explanation:

A+B+C=180°

55°+B+82°=180°

B=43°

Using the formulae

(Sin A)/a = (Sin B)/b

(Sin 55)/8 = (Sin 43)/b

b = [8(Sin 43)]/(Sin 55)

b= 6.7 cm

Nonamiya [84]3 years ago
4 0

Answer:

always b is equal to 9 is rhdx forum post in is ek of

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Given: Q = 7m + 3n, R = 11 - 2m, S = n + 5, and T = -m - 3n + 8.
Oxana [17]

Answer:

6m + n + 2

Step-by-step explanation:

[7m + 3n - 11 - 2m] + [n + 5 - -m - 3n + 8]

[5m + 3n - 11] + [n + 5 + m - 3n + 8]

5m + 3n - 11 + (-2n + 13 + m)

5m + 3n - 11 - 2n + 13 + m

6m + n + 2

3 0
3 years ago
Guys can u help me asap ?? i need explaination aloong the answer
Alisiya [41]

Answer:

5225 x 5/100 = 261.25

5225 - 261.25 = 4963.75

Step-by-step explanation:

We know that % means a 100 so if they say they allowed a 5% discount then that would be 5/100.

You will take the fraction and multiple it with the original price. It will then give you the discount price, you will then need to subtract the discounted amount from the original sale price which is 5225. Then you will get your final answer.

;-)

5 0
3 years ago
A particular psychological test is used to measure need for achievement. The average test score for all university students in O
Angelina_Jolie [31]

Answer:

Only B and C are always true.

Step-by-step explanation:

To analyse which statements are always true, we go through the process of finding confidence intervals for sample means, from the start.

Confidence Interval = (Sample mean) ± (Margin of error)

From this expression, it is evident that the margin of error determines how wide the confidence interval would be.

Sample Mean = 110 (given)

Margin of Error = (Critical value) × (Standard deviation of the distribution of sample means)

Since no information about the population standard deviation is provided, the critical value is obtained using t-distribution.

The critical value usually varies at different confidence levels and degree of freedoms.

The higher the confidence level, the higher the critical value and the higher the margin of error leading to a wider range.

Hence, a confidence interval of 95% will have a higher critical value than a confidence interval of 90%. Hence, statement C is proved once that 'for n = 100, the 95% confidence interval will be wider than the 90% confidence interval'.

After obtaining the critical value, we then obtain the standard deviation of the distribution of sample means or simply the standard error of the mean. This is given as

σₓ = σ/√n

where σ = standard deviation; which isn't given. The standard deviation might be high enough to guarantee that the Margin of error is high too for the confidence interval to contain 115 or low enough to ensure that the Margin of error is very small and the confidence interval will not contain 115.

Or the sample size might be high enough to make the standard error of the mean to be eventually small and lead to a small margin of error and the condidence interval will not contain 115.

The point is, it isn't always sure that the resulting interval.would contain 115. So, statement A isn't always true.

Then from σₓ = σ/√n,

n = sample size, a large sample size means a more narrow confidence interval and a small sample size means a wider sample size. This proves statement C.

The 95% confidence interval for n = 100 will be more narrow than the 95% confidence interval for n = 50.

Hence, Only B and C are always true.

Hope this Helps!!!

5 0
3 years ago
I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).
Aleksandr-060686 [28]

Answer:

\frac{7\pi}{24} and \frac{31\pi}{24}

Step-by-step explanation:

\sqrt{3} \tan(x-\frac{\pi}{8})-1=0

Let's first isolate the trig function.

Add 1 one on both sides:

\sqrt{3} \tan(x-\frac{\pi}{8})=1

Divide both sides by \sqrt{3}:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

Now recall \tan(u)=\frac{\sin(u)}{\cos(u)}.

\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}

or

\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}

The first ratio I have can be found using \frac{\pi}{6} in the first rotation of the unit circle.

The second ratio I have can be found using \frac{7\pi}{6} you can see this is on the same line as the \frac{\pi}{6} so you could write \frac{7\pi}{6} as \frac{\pi}{6}+\pi.

So this means the following:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

is true when x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

where n is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

Add \frac{\pi}{8} on both sides:

x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi

Find common denominator between the first two terms on the right.

That is 24.

x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi

x=\frac{7\pi}{24}+n \pi (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval [0,2\pi).

So if \sqrt{3} \tan(x-\frac{\pi}{8})-1=0 and we let u=x-\frac{\pi}{8}, then solving for x gives us:

u+\frac{\pi}{8}=x ( I just added \frac{\pi}{8} on both sides.)

So recall 0\le x.

Then 0 \le u+\frac{\pi}{8}.

Subtract \frac{\pi}{8} on both sides:

-\frac{\pi}{8}\le u

Simplify:

-\frac{\pi}{8}\le u

-\frac{\pi}{8}\le u

So we want to find solutions to:

\tan(u)=\frac{1}{\sqrt{3}} with the condition:

-\frac{\pi}{8}\le u

That's just at \frac{\pi}{6} and \frac{7\pi}{6}

So now adding \frac{\pi}{8} to both gives us the solutions to:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}} in the interval:

0\le x.

The solutions we are looking for are:

\frac{\pi}{6}+\frac{\pi}{8} and \frac{7\pi}{6}+\frac{\pi}{8}

Let's simplifying:

(\frac{1}{6}+\frac{1}{8})\pi and (\frac{7}{6}+\frac{1}{8})\pi

\frac{7}{24}\pi and \frac{31}{24}\pi

\frac{7\pi}{24} and \frac{31\pi}{24}

5 0
3 years ago
Please help this is my last thing to do!!!
Lana71 [14]

Answer:

area = 18x² + 63x + 55 square units

Second-degree trinomial.

Step-by-step explanation:

area = (3x+5)(6x+11) = 18x² + 63x + 55 square units

Second-degree trinomial.

5 0
3 years ago
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