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valentina_108 [34]

2 years ago

6

WILL MARK THE BRAINIEST PLZ HELP NOW!!! LOTS OF POINTS Determine how many solutions there are and if there are solutions solve f

or angle B. a) a=9, b=11 and A= 61 degrees b) a=25, b= 28 and A= 43 degrees.

1 answer:

harina [27]2 years ago

4 0

<h3>Answer:</h3>

a) no solutions

b) B = 49.8° or 130.2°

<h3>Explanation:</h3>

a) By the law of sines, ...

... sin(B)/b = sin(A)/a

... sin(B) = b·sin(A)/a = (11/9)·sin(61°) = 1.069 . . . . . not realizable

there is no solution

b) As before, ...

... sin(B) = (28/25)·sin(43°) ≈ 0.763838...

... B = arcsin(0.763838...) ≈ 49.80°

The side length b is longer than the side length a, so there will be two solutions. The other angle for B is the supplement to this one:

... B = 180° -49.8° = 130.2°

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An item costs x dollars and then a 20% discount is applied

Marat540 [252]

Answer:

If an item is x dollars and 20% discount is applied, the new price will be: 0.8x

Step-by-step explanation:

100 - 20%(20) = 80

80 = 80% or 0.8

So, the remaining of the price of x will be 0.8x.

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2 years ago

If 2a+4b=5 and a is equal to three times b, what is 3a?

IrinaK [193]

Answer:

3a=4.5

Step-by-step explanation:

it says that "a is equal to 3 times b", which translates to a=3b

We can substitute a as 3b in 2a+4b=5

2(3b)+4b=5

multiply

6b+4b=5

add

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divide by 10

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Hope this helps!

7 0

2 years ago

The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi

eimsori [14]

Answer:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=4.2$ represent the sample mean

$s=1.4$ represent the sample standard deviation

n=16 represent the sample selected

$\alpha=0.05$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:

Null hypothesis: $\mu = 3.6$

Alternative hypothesis: $\mu \neq 3.6$

If we analyze the size for the sample is < 30 and we know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-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:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with $df=n-1=16-1=15$ degrees of freedom. The value for $\alpha=0.05$ and $\alpha/2=0.025$ so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got $t_{crit}=\pm 2.131$

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

Conclusion

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

3 0

3 years ago

If the local linear approximation of f(x) = 2cos x + e^2x at x = 2 is used to find the approximation for f(2.1), then the % erro

yKpoI14uk [10]

<u>Answer-</u>

The % error of this approximation is 1.64%

<u>Solution-</u>

Here,

$\Rightarrow f(x) = 2\cos x + e^{2x}$

$\Rightarrow f'(x) = -2\sin x + 2e^{2x}$

And,

$\Rightarrow f(2) = 2\cos 2 + e^{4}$

$\Rightarrow f'(2) = -2\sin 2 + 2e^{4}$

Taking (2, f(2)) as a point and slope as, f'(2), the function would be,

$\Rightarrow y-y_1=m(x-x_1)$

$\Rightarrow y-(2\cos 2 + e^{4})=(-2\sin 2 + 2e^{4})(x-2)$

$\Rightarrow y=(-2\sin 2 + 2e^{4})(x-2)+(2\cos 2 + e^{4})$

The value of f(2.1) will be

$\Rightarrow y=(-2\sin 2 + 2e^{4})(2.1-2)+(2\cos 2 + e^{4})$

$\Rightarrow y=(-2\sin 2 + 2e^{4})(0.1)+(2\cos 2 + e^{4})$

$\Rightarrow y=64.5946$

According to given function, f(2.1) will be,

$\Rightarrow f(2.1) = 2\cos 2.1 + e^{2(2.1)}$

$\Rightarrow f(2.1) = 65.6766$

$\therefore \%\ error=\dfrac{65.6766-64.5946}{65.6766}=0.0164=1.64\%$

5 0

3 years ago

Complete the following statement.<br><br> 20% of $55 = $____

Eva8 [605]

Answer:

$11

Step-by-step explanation:

Let's take 20% and convert it into a decimal, which is .20. Multiply that percentage by $55 (.20 * 55) and you get $11, which is the answer.

3 0

2 years ago