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BaLLatris [955]
3 years ago
14

Hii i need to turn this assignment in but i need help

Mathematics
1 answer:
Troyanec [42]3 years ago
6 0

Answer:

1. 53

2. 43

Step-by-step explanation:

1. If angle A and B are supplementary that means they add up to 180 degrees. So 180-127= 53.

2. If angle A and B are complementary that means they add up to 90 degrees. So 90-47= 43.

You might be interested in
Trigonometry
xxMikexx [17]

Answer:

1.309 rad

Step-by-step explanation:

We are told to convert 75° to radians.

# we know that 1° is equivalent to 0.01745radians.

-To convert any angle degree to  radians, we multiply it by π/180 or 0.01745rad:

=75\textdegree \times \pi/180\\\\=1.309 \ rad

Hence, the radian equivalent of 75 is 1.309 rad

8 0
3 years ago
(-y+5•3)+7.2y-9)=6.2y+n
kramer
(-y+5x3)+(7.2y-9)=6.2y+n
(-y+15)+(7.2y-9)=6.2y+n
since you're adding the two parentheses, you don't need to have them there
-y+15+7.2y-9=6.2y+n
7.2y-y +15-9 =6.2y+n
6.2y + 6 =6.2y+n
6.2y - 6.2y -n = -6
-n=-6
n=6

3 0
3 years ago
Please answer all! I appreciate it
irga5000 [103]

1.7(a+3)

2.5(x-1)

3.prime

4.3(5p-9)

5.3(15k-4)

6.8(3p+1)

7.47-5-2m+8m=42+6m=6(7+m)

8.2*10a-2*7b-10b-2a

20a-14b-10b-2a

18a-24b=6(3a-4b)

4 0
3 years ago
The age of United States Presidents on the day of their first inauguration follows a Normal distribution with mean 56 and standa
Talja [164]

Answer:

a) 0.7088 = 70.88% probability that a randomly selected President was less than 60 years old on the day of their first inauguration.

b) The 75th percentile for the age of United States Presidents on the day of inauguration is 61.

c) 0.8643 = 86.43% probability that the average age on the day of their first inauguration for a random sample of 4 United States Presidents exceeds 60 years.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The age of United States Presidents on the day of their first inauguration follows a Normal distribution with mean 56 and standard deviation 7.3.

This means that \mu = 56, \sigma = 7.3

(a) (5 points) Compute the probability that a randomly selected President was less than 60 years old on the day of their first inauguration.

This is the pvalue of Z when X = 60. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{60 - 56}{7.3}

Z = 0.55

Z = 0.55 has a pvalue of 0.7088

0.7088 = 70.88% probability that a randomly selected President was less than 60 years old on the day of their first inauguration.

(b) (5 points) Compute the 75th percentile for the age of United States Presidents on the day of inauguration.

This is X when Z has a pvalue of 0.75. So X when Z = 0.675.

Z = \frac{X - \mu}{\sigma}

0.675 = \frac{X - 56}{7.3}

X - 56 = 0.675*7.3

X = 61

The 75th percentile for the age of United States Presidents on the day of inauguration is 61.

(c) (5 points) Compute the probability that the average age on the day of their first inauguration for a random sample of 4 United States Presidents exceeds 60 years.

Now, by the Central Limit Theorem, we have that n = 4, s = \frac{7.3}{\sqrt{4}} = 3.65

This is the pvalue of Z when X = 60. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{60 - 56}{3.65}

Z = 1.1

Z = 1.1 has a pvalue of 0.8643

0.8643 = 86.43% probability that the average age on the day of their first inauguration for a random sample of 4 United States Presidents exceeds 60 years.

4 0
3 years ago
How we supposed to find the answer?
Gnoma [55]

Answer:

5^6

Step-by-step explanation:

(5^3)^2

We know that a^b^c = a^(b*c)

5^(3*2)

5^6

8 0
3 years ago
Read 2 more answers
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