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2 years ago

Find the next two numbers in the sequence 5,2,-1,…

Mathematics

1 answer:

Nataly_w [17]2 years ago

5 0

Answer:

The answer is (-4) and (-7).

Step-by-step explanation:

<h3><u>Given</u>;</h3>

a₁ = 5
a₂ = 2
a₃ = (-1)

a₄ = ?
a₅ = ?

<h3><u>Formula</u>;</h3>

aₙ = a + (n – 1) × d

Now, for common difference (d)

d = a₂ – a₁ = 2 – 5 = -3

d = a₃ – a₂ = (-1) – 2 = -3

Here, common difference is same everywhere.

So, for fourth term or n = 4

aₙ = a + (n – 1) × d

a₄ = 5 + (4 – 1) × (-3)

a₄ = 5 + 3 × (-3)

a₄ = 5 – 9

a₄ = -4

Then, for fifth term or n = 5

a₅ = 5 + (5 – 1) × (-3)

a₅ = 5 + 4 × (-3)

a₅ = 5 – 12

a₅ = -7

Thus, The fourth term is (-4) and fifth term is (-7).

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The metal in a penny is worth about 0.505 cent. How many cents is this written as a fraction

valentinak56 [21]

It would be 505/1000 but If you need it to be reduced it would be 101/200

4 0

3 years ago

The Alvin Secretarial Service procures temporary office personnel for major corporations. They have found that 60% of their invo

Hoochie [10]

Answer:

0.9942 = 99.42% probability that at least 6 of the invoices will be paid within ten working days

Step-by-step explanation:

For each invoice, there are only two possible outcomes. Either they are paid within 10 working days, or they are not. The probability of an invoice being paid within 10 working days is independent of other invoices. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

60% of their invoices are paid within ten working days.

This means that $p = 0.6$

A random sample of 18 invoices is checked.

This means that $n = 18$

What is the probability that at least 6 of the invoices will be paid within ten working days?

Either less than 6 are paid within 10 working days, or at least 6 are paid. The sum of the probabilities of these events is 1. So

$P(X < 6) + P(X \geq 6) = 1$

We want $P(X \geq 6)$ . So

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

Then

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{18,0}.(0.6)^{0}.(0.4)^{18} \approx 0$

$P(X = 1) = C_{18,1}.(0.6)^{1}.(0.4)^{17} \approx 0$

$P(X = 2) = C_{18,2}.(0.6)^{2}.(0.4)^{16} \approx 0$

$P(X = 3) = C_{18,3}.(0.6)^{3}.(0.4)^{15} = 0.0002$

$P(X = 4) = C_{18,4}.(0.6)^{4}.(0.4)^{14} = 0.0011$

$P(X = 5) = C_{18,5}.(0.6)^{5}.(0.4)^{13} = 0.0045$

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0 + 0 + 0 + 0.0002 + 0.0011 + 0.0045 = 0.0058$

Finally

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.0058 = 0.9942$

0.9942 = 99.42% probability that at least 6 of the invoices will be paid within ten working days

6 0

4 years ago

a waiter had nine tables he was waiting on, with seven women and three men at each table. how many customees total did the waite

Greeley [361]

Since each table had 7 women and 3 men at each table, that would mean each table has 10 people. Since there are 9 tables he was waiting on, and 10 people at each table, multiply 9 x 10, and you will find that he served 90 people in total.

8 0

4 years ago

Read 2 more answers

Samantha has 7/3 cups of ice cream to make milk shake.Each shake uses 1/8 cup of ice cream. How many milk shakes can she make?

Travka [436]

18 milkshakes
Hope this helps

7 0

3 years ago

Will the quotient of 2÷1/3 be smaller or bigger than each term? Explain in words.

RSB [31]

In short, if the denominator is a decimal or a fraction and you have an integer in the numerator, the numerator gets blown up.

so if you divide by 1/2, the numerator blows up to twice as large, if you divide by 1/5, to five times as large, if you divide by 1/3... well, let's see,

$\bf 2\div \cfrac{1}{3}\implies \cfrac{2}{1}\div \cfrac{1}{3}\implies \cfrac{2}{1}\cdot \cfrac{3}{1}\implies \cfrac{6}{1}\implies 6$

why is that? well, you're dividing, and so is saying "how many times 1/3 goes into 2"?

well, 1/3 + 1/3 + 1/3 = 3/3 or 1, so in 1 whole, the fraction goes there 3 times, so in 2 wholes, it'll be 6 times.

3 0

3 years ago