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2 years ago

Please help ASAP!!!!!!!

Mathematics

1 answer:

VLD [36.1K]2 years ago

3 0

Answer:

10th term is 10

Step-by-step explanation:

The nth term for finding the geometric progression is expressed as;

Tn = ar^n-1

a is the first term

r is the common ratio

n is the number of terms

a11 = ar^11-1

a11 = ar^10

Since a11 = -5 and r = -1/2

-5 = a(-1/2)^10

-5 = a(1/1024)

a= 1024 * -5

a = -5120

Nest is to get the 10th terms

a10 = ar^9

a10 = -5120 * (-1/2)^9

a10 = -5120 * -1/512

a10 = 10

Hence the 10th term of the sequence is 10

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Find using appropriate Properties:<br> 6/7 x -2/3 + 3/5 + 5/8 x -6/7

Valentin [98]

The simplified form of the expression is given as 83x/56 -(97/105)

<h3>Finding the value of unknown variables</h3>

Alphabets are usually represented as unknown variables in a equation or expression.

Given the following expression shown below

6/7 x -2/3 + 3/5 + 5/8 x -6/7

Collect the like terms to have;

6/7 x+ 5/8 x -2/3 + 3/5 -6/7

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83x/56 -(97/105)

Hence the simplified form of the expression is given as 83x/56 -(97/105)

Learn more on expression simplification here: brainly.com/question/723406

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1 year ago

The average weekly salary of two employees is $5350. One makes $250 more than the other. Find their salaries.

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3 years ago

Evaluate the function f(x) = -4x + 2, when x = -1.

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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.

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3 years ago