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Studentka2010 [4]

2 years ago

5

48 people are at a wedding reception. The tables are arranged so that the same number of people sit at each table. How many peop

le sit at each table?
Give two possible answers
(I do know the answer but I just wanna test you guys)

1 answer:

geniusboy [140]2 years ago

6 0

Answer:

Step-by-step explanation:

Simply if there are 6 tables so 48 ÷ 6 = 8 people on each table

So it depends on the number of table

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It cost Jayden $4.80 to send 96 text messages. How much would it cost to send 155 text messages?

TEA [102]

96 messages .... $4.80
155 messages .... x

x= (155x4.80):96=7.75

3 0

2 years ago

ALGEBRA 2/ TRIG QUESTION ATTACHED Please help ❤️❤️

ss7ja [257]

Let's calculate the value of angle A and B

sin(A) =-4/5 → sin⁻¹(- 4/5) = A → A = - 53.13

cos(B) = -5/13 → cos⁻¹ (- 5/13) = B → B = 112.62

tan (A+B) = sin(A+B)/cos(A+B) with A+B = -53.13 + 112.62 = 59.49

tan (A+B) = sin(59.49)/cos(59.49) = 0.86154/0.507688 = 1.6969.

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2 years ago

Write an arithmetic series for which S5 = 10.

Ludmilka [50]

S5=10 means the 5th number is 10.
So the arithmetic sequence is s1, s2,s3, s4, 10

3 0

2 years ago

It is claimed that 55% of marriages in the state of California end in divorce within the first 15 years. A large study was start

kykrilka [37]

Answer:

0.0045 = 0.45% probability that less than two of them ended in a divorce

Step-by-step explanation:

For each marriage, there are only two possible outcomes. Either it ended in divorce, or it did not. The probability of a marriage ending in divorce is independent of any other marriage. This means that the binomial probability distribution is used 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.

55% of marriages in the state of California end in divorce within the first 15 years.

This means that $p = 0.55$

Suppose 10 marriages are randomly selected.

This means that $n = 10$

What is the probability that less than two of them ended in a divorce?

This is

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

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

$P(X = 0) = C_{10,0}.(0.55)^{0}.(0.45)^{10} = 0.0003$

$P(X = 1) = C_{10,1}.(0.55)^{1}.(0.45)^{9} = 0.0042$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0003 + 0.0042 = 0.0045$

0.0045 = 0.45% probability that less than two of them ended in a divorce

8 0

1 year ago

An employment agency requires applicants average at least 70% on a battery of four job skills tests. If an applicant scored 70%,

brilliants [131]

Answer:

atleast 52

Step-by-step explanation:

Given that an employment agency requires applicants average at least 70% on a battery of four job skills tests.

An applicant scored 70%, 77%, and 81% on the first three exams,

Since weightages are not given we can assume all exams have equal weights

Let x be the score on the 4th test

Then total of all 4 exams = $70+77+81+x\\= 228+x$

Average should exceed 70%

i.e. $\bar X \geq 70\\Total\geq 70(4) =280$

Comparing the two totals we have

$228+x\geq 280\\x\geq 280-228 = 52$

Hemust score on the fourth test a score atleast 52 to maintain a 70% or better average.

3 0

2 years ago