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2 years ago

A triangle has the following measurements: B=79 degrees a=5ft

1 answer:

6 0

The law of cosines for this particular situation is b^2 = a^2 + c^2 - 2ac cosB.
Filling in what you know, you have b^2 = 25 + 49 - [2(5)(7)-.1908], which simplifies to b^2 = 74 - 69.8092 which gives you a b^2 value of 4.1908, but you have to take the square root of that so you get a side value for b of 2.05.

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A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day of sharpness. If any dull blade

Alex73 [517]

Answer:

Part a

The probability that assembly is replaced the first day is 0.7069.

Part b

The probability that assembly is replaced no replaced until the third day of evaluation is 0.0607.

Part c

The probability that the assembly is not replaced until the third day of evaluation is 0.2811.

Step-by-step explanation:

Hypergeometric Distribution: A random variable x that represents number of success of the n trails without replacement and M represents number of success of the N trails without replacement is termed as the hypergeometric distribution. Moreover, it consists of fixed number of trails and also the two possible outcomes for each trail.

It occurs when there is finite population and samples are taken without replacement.

The probability distribution of the hyper geometric is,

$P(x,N,n,M)=\frac{(\limits^M_x)(\imits^{N-M}_{n-x})}{(\limits^N_n)}$

Here x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

Probability that at least one of the trail is succeed is,

$P(x\geq1)=1-P(x$

(a)

Compute the probability that the assembly is replaced the first day.

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48.

Number of blades selected at random from the assembly n= 5

Number of blades in an assembly dull is M = 10.

The probability mass function is,

$P(X=x)=\frac{[\limits^M_x][\limits^{N-M}_{n-x}]}{[\limits^N_n]};x=0,1,2,...,n\\\\=\frac{[\limits^{10}_x][\limits^{48-10}_{5-x}]}{[\limits^{48}_5]}$

The probability that assembly is replaced the first day means the probability that at least one blade is dull is,

$P(x\geq 1)=1- P(x$

(b)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly N = 5

Number of blades in an assembly dull is M = 10

From the information,

The probability that assembly is replaced (P) is 0.7069.

The probability that assembly is not replaced is (Q) is,

$q=1-p\\= 1-0.7069= 0.2931$

The geometric probability mass function is,

$P(X = x)= q^{x-1} p; x =1,2,....=(0.2931)^{x-1}(0.7069)$

The probability that assembly is replaced no replaced until the third day of evaluation is,

$P(X = 3)=(0.2931)^{3-1}(0.7069)\\=(0.2931)^2(0.7069)= 0.0607$

(c)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly n = 5

Suppose that on the first day of the evaluation two of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 2.

$P(x=0)=\frac{(\limits^2_0)(\limits^{48-2}_{5-0})}{\limits^{48}_5}\\\\=\frac{(\limits^{46}_5)}{(\limits^{48}_5)}\\\\= 0.8005$

Suppose that on the second day of the evaluation six of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 6.

$P(x=0)=\frac{(\limits^6_0)(\limits^{48-6}_{5-0})}{(\limits^{48}_5)}\\\\=\frac{(\limits^{42}_5}{(\limits^{48}_5)}\\\\= 0.4968$

Suppose that on the third day of the evaluation ten of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M

= 10.

$P(x\geq 1)=1- P(x$

The probability that the assembly is not replaced until the third day of evaluation is,

P(The assembly is not replaced until the third day)=P(The assembly is not replaced first day) x P(The assembly is not replaced second day) x P(The assembly is replaced third day)

=(0.8005)(0.4968)(0.7069)= 0.2811

5 0

3 years ago

YOOOOOOOO I FINALLY WAS ABLE TO GET REGISTERED

Agata [3.3K]

Answer:

Ty for the free points lol

8 0

3 years ago

Read 2 more answers

Paisley has 8\tfrac{1}{4}8 4 1 cups of yogurt to make smoothies. Each smoothie uses \tfrac{11}{16} 16 11 cup of yogurt. How

Ivenika [448]

Given:

Total Yogurt = $8\dfrac{1}{4}$ cups

Yogurt required for each smoothie = $\dfrac{11}{16}$ cup

To find:

The number of smoothies that Paisley can make with the yogurt.

Solution:

We know that,

$\text{Number of smoothies}=\dfrac{\text{Total yogurt}}{\text{Yogurt required for each smoothie}}$

Substituting the given values, we get

$\text{Number of smoothies}=\dfrac{8\dfrac{1}{4}}{\dfrac{11}{16}}$

$\text{Number of smoothies}=8\dfrac{1}{4}\times \dfrac{16}{11}$

$\text{Number of smoothies}=\dfrac{32+1}{4}\times \dfrac{16}{11}$

$\text{Number of smoothies}=\dfrac{33}{4}\times \dfrac{16}{11}$

$\text{Number of smoothies}=3\times 4$

$\text{Number of smoothies}=12$

Therefore, the number of smoothies is 12.

6 0

3 years ago

Vernon tossed a coin 20 times. The results were 8 head s and 12 tails. What is the experimental probability of tossing heads?

solmaris [256]

Answer:

2/5

Step-by-step explanation:

The experimental probability is

P (heads) = number of heads/ total tosses

= 8/20

= 2/5

8 0

3 years ago

So is anyone good at geometry ive posted this before but i was used for points i need help

SOVA2 [1]

Answers:
Reason 3: Definition of Parallelogram
Reason 4: Alternate Interior Angles Theorem
Reason 5: Reflexive Property of Congruence
Reason 6: ASA Congruence Property

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Explanations:

Explanation for Reason 3: A parallelogram, by definition, has opposite sides that are parallel. It's built into the name more or less. Sides AB and CD are opposite one another in the parallelogram so they are parallel segments

Explanation for Reason 4: Angle ABD is congruent to angle CDB because they are alternate interior angles. They are on the inside of the "train tracks" that are formed by AB and CD. They lay on opposite sides of the transversal BD

Explanation for Reason 5: Any segment is congruent to itself; ie, the same length

Explanation for Reason 6: Using reasons 2,5 and 4, we can use ASA (angle side angle) to prove the two triangles ABD and CDB congruent. Reason 2 is the first "A" in ASA. Reason 5 is the S in ASA. Reason 4 is the other A in ASA. The side is between the two pairs of angles. See the attache image for a visual summary of how ASA is being used.

8 0

2 years ago