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2 years ago

Find m •23 •43 •65 •67

Mathematics

2 answers:

mina [271]2 years ago

4 0

Cos x = adjacent/hypotenuse
Cos x = 5/13
MM

nika2105 [10]2 years ago

4 0

Cos F=adjacent/hypotenuse
cos F=5/13
F=arccos(0.3846)
F=~67 degrees

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Which of the fractions below are less than 2/5? Select two.

My name is Ann [436]

Answer:

1/8 is less than

Step-by-step explanation:

i dont see any fractions below gona have to edit your answer

7 0

3 years ago

The sum of four consecutive odd integers is three more than five times the least of the integers. Find the integers.

Olegator [25]

Answer:

Step-by-step explanation:

{k - some integer}

2k+1 - the first odd integer (the least)

5(2k+1) - five times the least

5(2k+1)+3 - three more than five times the least

2k+1+2 = 2k+3 - the odd integer consecutive to 2k+1

2k+3+2 = 2k+5 - the next odd consecutive integer (third)

2k+5+2 = 2k+7 - the last odd consecutive integer (fourth)

2k+1+2k+3+2k+5+2k+7 - the sum of four odd consecutive integers

2k+1 + 2k+3 + 2k+5 + 2k+7 = 5(2k+1) + 3

8k + 16 = 10k + 5 + 3

- 10k -10k

-2k + 16 = 8

-16 - 16

-2k = -8

÷(-2) ÷(-2)

k = 4

2k+1 = 2•4+1 = 9

2k+3 = 2•4+3 = 11

2k+5 = 2•4+5 = 13

2k+7 = 2•4+7 = 15

Check: 9+11+13+15 = 48; 48-3 = 45; 45:5 = 9 = 2k+1

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3 years ago

Which line is parallel to line r? line p line q line s line t

Mila [183]

Line Q??? You don't really have an image of it

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3 years ago

Read 2 more answers

The st. joe company grows pine trees and the average annual increase in tree diameter is 3.1 inches with a standard deviation of

77julia77 [94]

Solution: We are given:

μ=3.1,σ=0.5,n=50

We have to find P(Mean <2.9)

We need to first find the z score

z= (xbar-μ)/(σ/sqrt(n))

=(2.9-3.1)/(0.5/sqrt(50))

=(-0.2)/0.0707

=-2.83

Now we have to find P(z<-2.83)

Using the standard normal table, we have:

P(z<-2.83)=0.0023

Therefore the probability of the sample mean being less the 2.9 inches is 0.0023

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2 years ago

One household is to be selected at random from a town. The probability that the household has a cat is 0.20.2 . The proba

dimulka [17.4K]

Answer:

There is a 50% probability that the household has a dog, given that the household has a cat.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a household has a cat.

B is the probability that a household has a dog.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a household has a cat but not a dog and $A \cap B$ is the probability that a household has both a cat and a dog.

By the same logic, we have that:

$B = b + (A \cap B)$

The probability that the household has a cat or a dog is 0.5

$a + b + (A \cap B) = 0.5$

The probability that the household has a dog is 0.4

$B = 0.4$

$B = b + (A \cap B)$

$b = 0.4 - (A \cap B)$

The probability that the household has a cat is 0.2.

$A = 0.2$

$A = a + (A \cap B)$

$a = 0.2 - (A \cap B)$

$a + b + (A \cap B) = 0.5$

$0.2 - (A \cap B) + 0.4 - (A \cap B) + (A \cap B) = 0.5$

$A \cap B = 0.1$

What is the probability that the household has a dog, given that the household has a cat?

20% of the households have a cat, and 10% have both a cat and a dog. So

$P = \frac{A \cap B}{A} = {0.1}{0.2} = 0.5$

There is a 50% probability that the household has a dog, given that the household has a cat.

4 0

2 years ago