All categories

3 years ago

4, 13, 22, ... Find the 35th term.

Mathematics

2 answers:

Inessa [10]3 years ago

8 0

Answer:

im thinking its 40.. tho im not sure im pretty sure it is so im sorry if its wrong

Step-by-step explanation:

Mice21 [21]3 years ago

5 0

Answer:

310

Step-by-step explanation:

1.Find your constant difference of the sequence by subtracting from left to right which is 9

2. find your fomula by using Tn=bn+C (b is your constant difference you can find C by C=T1 -b ) which will then give us 9n-5

3. T(35)=9n-5=9(35)-5=310

You might be interested in

A survey is conducted to find out whether people in metropolitan areas obtain their news from television (Event T), an newspaper

MrMuchimi

Answer:

e) P ( R' | T ) = 0.62857

f) P ( T' | N ) = 0.32258

g) P ( R' | N ) = 0.70968

h) P ( T' | N&R ) = 5/6

Step-by-step explanation:

Given:

- The probability of Television as news source P ( T ) = 0.7

- The probability of Newspaper as news source P ( N ) = 0.62

- The probability of Radio as news source P ( R ) = 0.46

- The probability of Television & Newspaper as news source P (T&N) = 0.42

- The probability of Television & Radio as news source P (T&R) = 0.26

- The probability of Radio & Newspaper as news source P (R&N) = 0.18

- The probability of all 3 as news source P ( T & R & N ) = 0.03

Find:

Given that TV is a news source, what is the probability that radio is not a news source?

Given that newspaper is a news source, what is the probability that TV is not a news source?

Given that newspaper is a news source, what is the probability that radio is not a news source?

Given that both newspaper and radio are news sources, what is the probability that TV is not a news source?

Solution:

- We will first compute the individual probability of each event occurring alone.

P (Television is the only news source) = P(T) - P(T&R) - P(T&N) + P(T&N&R)

P ( Only Television ) = P ( only T ) = 0.7 - 0.42 - 0.26 + 0.03 = 0.05

P (Newspaper is the only news source) = P(N)-P(N&R)-P(T&N)+P(T&N&R)

P ( Only Newspaper ) = P ( only N ) = 0.62 - 0.18 - 0.42 + 0.03 = 0.05

P (Radio is the only news source) = P(R) - P(T&R) - P(N&R) + P(T&N&R)

P ( Only Radio ) = P ( only R ) = 0.46 - 0.26 - 0.18 + 0.03 = 0.05

- Now for part e)

We are asked for a conditional probability of the form as follows:

P ( R' | T ) = P ( R' & T ) / P ( T )

First compute the probability that next news source is not Radio provided it is already a source of TV.

P ( R' & T ) = P( only T ) + P ( only T & N ) = 0.05 + 0.42 - 0.03 = 0.44

Hence,

P ( R' | T ) = 0.44 / 0.7

P ( R' | T ) = 0.62857

- Now for part f)

We are asked for a conditional probability of the form as follows:

P ( T' | N ) = P ( T' & N ) / P ( N )

First compute the probability that next news source is not TV provided it is already a source of Newspaper.

P ( T' & N ) = P( only N ) + P ( only R & N ) = 0.05 + 0.18 - 0.03 = 0.2

Hence,

P ( T' | N ) = 0.2 / 0.62

P ( T' | N ) = 0.32258

- Now for part g)

We are asked for a conditional probability of the form as follows:

P ( R' | N ) = P ( R' & N ) / P ( N )

First compute the probability that next news source is not Radio provided it is already a source of Newspaper.

P ( R' & N ) = P( only N ) + P ( only T & N ) = 0.05 + 0.42 - 0.03 = 0.44

Hence,

P ( R' | N ) = 0.44 / 0.62

P ( R' | N ) = 0.70968

- Now for part h)

We are asked for a conditional probability of the form as follows:

P ( T' | N&R ) = P ( T' & N & R) / P ( N & R )

First compute the probability that next news source is not TV provided it is already a source of both radio and Newspaper.

P ( T' & N & R) = P ( only N & R ) = 0.18 - 0.03 = 0.15

Hence,

P ( T' | N&R ) = 0.15 / 0.18

P ( T' | N&R ) = 5/6

6 0

3 years ago

sarah runs a day camp in the summer and needs to buy paper towels and soap for the restrooms. She spends $129 on t cases of pape

Fofino [41]

Pt 7 × 12=84 and soap 15×3 =45 add 84+45=129

4 0

4 years ago

What is the result when the number 90 is increased by 10%?

AleksandrR [38]

90 * .1 = 9
90 +9 = 99

7 0

3 years ago

Read 2 more answers

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05