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1 year ago

14

You're given three angle measurements of 30 degrees, 70 degrees, and 80 degrees. How many triangles can you construct using thes

e measurements?

1 answer:

Strike441 [17]1 year ago

7 0

180-degree triangle so practically 1 triangle but i am not 100% sure

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At a carnival, an individual can win a prize by choosing a rubber duck from a pond with "Win" written on the underside of the du

boyakko [2]

Step-by-step explanation:

The probability of success = 8/(8 + 17) = 8/25 = 0.32.

Let X be the random variable denoting the number of successes (number of times the individual won a prize) in four picks.

Hence, X ~ Bin(4, 0.32).

Thus, P(X = 1) = $4C_1=(0.32)(1-0.68)^{4-1}=4C_1(0.32)(0.68)^3$

3 0

1 year ago

Two con terminal angles 3pi/4 negative and positive answer in radians

son4ous [18]

Answer:

Negative Coterminal: -5π/4

Positive Coterminal: 11π/4

Step-by-step explanation:

The easiest way to find <em>specific </em>(not infinite) coterminal values is to ±2π. When you subtract 2π, you will get a negative coterminal. When you add 2π, you will get a positive coterminal. Keep in mind though that a tan∅ or cot∅ only needs ±π, not ±2π.

8 0

1 year ago

For the equation ae^ct=d, solve for the variable t in terms of a,c, and d. Express your answer in terms of the natural logarithm

saveliy_v [14]

We have been given an equation $ae^{ct}=d$ . We are asked to solve the equation for t.

First of all, we will divide both sides of equation by a.

$\frac{ae^{ct}}{a}=\frac{d}{a}$

$e^{ct}=\frac{d}{a}$

Now we will take natural log on both sides.

$\text{ln}(e^{ct})=\text{ln}(\frac{d}{a})$

Using natural log property $\text{ln}(a^b)=b\cdot \text{ln}(a)$ , we will get:

$ct\cdot \text{ln}(e)=\text{ln}(\frac{d}{a})$

We know that $\text{ln}(e)=1$ , so we will get:

$ct\cdot 1=\text{ln}(\frac{d}{a})$

$ct=\text{ln}(\frac{d}{a})$

Now we will divide both sides by c as:

$\frac{ct}{c}=\frac{\text{ln}(\frac{d}{a})}{c}$

$t=\frac{\text{ln}(\frac{d}{a})}{c}$

Therefore, our solution would be $t=\frac{\text{ln}(\frac{d}{a})}{c}$ .

5 0

2 years ago

Suppose the time required for an auto shop to do a tune-up is normally distributed, with a mean of 102 minutes and a standard de

hammer [34]

Answer:

Step-by-step explanation:

Suppose the time required for an auto shop to do a tune-up is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - u)/s

Where

x = points scored by students

u = mean time

s = standard deviation

From the information given,

u = 102 minutes

s = 18 minutes

1) We want to find the probability that a tune-up will take more than 2hrs. It is expressed as

P(x > 120 minutes) = 1 - P(x ≤ 120)

For x = 120

z = (120 - 102)/18 = 1

Looking at the normal distribution table, the probability corresponding to the z score is 0.8413

P(x > 120) = 1 - 0.8413 = 0.1587

2) We want to find the probability that a tune-up will take lesser than 66 minutes. It is expressed as

P(x < 66 minutes)

For x = 66

z = (66 - 102)/18 = - 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.02275

P(x < 66 minutes) = 0.02275

4 0

2 years ago

A psychologist is collecting data on the time it takes to learn a certain task. For 50 randomly selected adult subjects, the sam

tatuchka [14]

Answer: $(15.263,\ 17.537)$

Step-by-step explanation:

According to the given information, we have

Sample size : n= 50

$\overline{x}=16.40$

$s=4.00$

Since population standard deviation is unknown, so we use t-test.

Critical value for 95 percent confidence interval :

$t_{n-1,\alpha/2}=t_{49, 0.025}= 2.009575\approx2.010$

Confidence interval : $\overline{x}\pm t_{n-1, \alpha/2}\dfrac{s}{\sqrt{n}}$

$16.40\pm (2.010)\dfrac{4}{\sqrt{50}}\\\\=16.40\pm1.13702770415\\\\=16.40\pm1.1370\\\\=(16.40-1.1370,\ 16.40+1.1370)\\\\=(15.263,\ 17.537)$

Required 95% confidence interval : $(15.263,\ 17.537)$

8 0

1 year ago