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SVEN [57.7K]
2 years ago
8

Find the nth term of this 5 7 9 11 13

Mathematics
2 answers:
denpristay [2]2 years ago
8 0
5 7 9 11 13 15 17 19 21

Hoped this helped
NemiM [27]2 years ago
4 0
Firstly, seeing that the sequence is based on odd numbers I would say that this is the formula (which will allow you to know whatever term):

nth term = 2n-1

Hope this helps!
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Runners in the half marathon entered in three waves. There were 341 runners in the first wave, 295 runners in the second wave, a
Natasha_Volkova [10]
190 runners would not have finished the race.

If you add up all the waves, there were 952 total runners.

If 7/8 finished the race, then that means that 1/8 did not finish the race.

Just multiply 1/8 by 952 to get the amount that did not finish.
It would be 190.
6 0
2 years ago
Which binomials below is a factor of this trinomial 5x^2-5x-100
34kurt
We are asked to determine the correct factorization of the given trinomial which is 5x² - 5x -100. The solution to this problem is shown below:
5x² - 5x - 100 let the problem equal to zero such as:
5x² - 5x - 100 = 0 , factor out 5 from the given trinomial
5(x²-x-20) = 0
5 (x-5)(x+4) = 0

The correct factor is 5(x-5)(x+4).
3 0
3 years ago
Day care centers expose children to a wider variety of germs than the children would be exposed to if they stayed at home more o
Ostrovityanka [42]

Answer:

a) The study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is approximately 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is approximately 0.8565

c) The odds ratio is approximately 0.6780

d) The 95% CI is approximately 0.523 < OR < 0.833

e) i) Yes

ii) Children with more social activity have a higher occurrence of acute lymphoblastic leukemia

Step-by-step explanation:

a) An experimental study is a study in which the researcher adds inputs to the study and then monitors the outcomes

An observational study is one in which the researcher observes risk factors and does not intervene in the process

Therefore, the study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is \hat p_1 = 1020/1272 = 85/106 = 0.8\overline {0188679245283} ≈ 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is \hat p_2 = 5343/6238 ≈ 0.8565

c) We have the following two way table;

\begin{array}{ccc}Lymphoblastic \ Leukemia &With \ Social \ Activity & Without \ Social \ Activity\\With&1020 \ (a)&252 \ (b)\\Without&5343 \ (c)&895 \ (d) \end{array}

The \ odds \ ratio = \dfrac{1,020 \times 895}{5,343 \times 252} \approx 0.6780

The odds ratio ≈ 0.6780

d) The 95% confidence interval for the odds ratio is given as follows;

95 \% \ CI = OR \ \pm \ 1.96 \times \sqrt{\dfrac{1}{a} +\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}}

Where;

a = 1020, b = 252, c = 5343, d = 895

Therefore;

95 \% \ CI \approx  0.6780 \ \pm \ 1.96 \times \sqrt{\dfrac{1}{1,020} +\dfrac{1}{252}+\dfrac{1}{5,343}+\dfrac{1}{895}} \approx 0.678 \pm0.155

The 95% CI ≈ 0.523 < OR < 0.833

e) i) Given that the observed Odds Ratio is within the Confidence Interval, therefore, there is an indication that the amount of social activity is associated with acute lymphoblastic leukemia

ii) Based on the proportion of the study findings, children with more social activity have a higher occurrence of acute lymphoblastic leukemia

6 0
2 years ago
The weights of 1250 pumpkins are normally distributed with mean of 28 pounds and a standard deviation of 6 pounds. About how man
Mekhanik [1.2K]
Kropot72
kropot72 3 years ago
This can be solved by using a standard normal distribution table. The z-score for 34 pounds is 1, the reason being that 34 is one standard deviation above the mean of 28 pounds.
Can use the table to find the cumulative probability for z = 1.00 and post the result? If you do this we can do the next simple steps.
5 0
2 years ago
The sum of 2 positive integers is 3. the sum of their squares is 5. find the 2 numbers
Kipish [7]

Answer:

2 and 1

Step-by-step explanation:

1 + 2 = 3

(1*1) + (2*2) = 5

5 0
3 years ago
Read 2 more answers
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