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mars1129 [50]
3 years ago
15

A markdown that is less than 1%

Mathematics
1 answer:
Alexxx [7]3 years ago
8 0
Then mark it down by less then 1%, of by whatever the number would be. 
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To describe a specific arithmetic sequence, Elijah wrote the recursive formula:
Rufina [12.5K]

Answer:

Y(n) = 7n + 23

Step-by-step explanation:

Given:

f(0) = 30

f(n+1) = f(n) + 7

For n=0 : f(1) = f(0) + 7

For n=1 :  f(2) = f(1) + 7

For n=2 : f(3) = f(2) + 7 and so on.

Hence the sequence is an arithmetic progression with common difference 7 and first term 30.

We have to find a general equation representing the terms of the sequence.

General term of an arithmetic progression is:

T(n) = a + (n-1)d

Here a = 30 and d = 7

Y(n) = 30 + 7(n-1) = 7n + 23

7 0
3 years ago
Estephany has a box that is filled with toys. The box has a length of 3.5 feet, a width of 6 feet, and a height of 3 feet. What
Zigmanuir [339]

Answer:

63

Step-by-step explanation:

The formula for volume of a box is length×width×height = volume

So you would multiply  3.5×6×3=63

6 0
3 years ago
Bryce has to complete 100 minutes of writing every ten days. He writes for 10 minutes each day. What fraction of his writing req
Sergio [31]
Since he completes 10 minutes each day for five days, he completes 50 minutes.

The percentage of his total requirement would be 50/100 = 0.5 = 50%.
5 0
3 years ago
Read 2 more answers
A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In
lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

                        P.Q. = \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }  ~ N(0,1)

where, \hat p_1 = sample proportion of males having blood disorder= \frac{250}{1000} = 0.25

\hat p_2 = sample proportion of females having blood disorder = \frac{275}{1000} = 0.275

n_1 = sample of males = 1000

n_2 = sample of females = 1000

p_1 = population proportion of males having blood disorder

p_2 = population proportion of females having blood disorder

<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u>p_1-p_2<u>)</u><u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                             of significance are -1.96 & 1.96}  

P(-1.96 < \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < {(\hat p_1-\hat p_2)-(p_1-p_2)} < 1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

P( (\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < (p_1-p_2) < (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

<u>95% confidence interval for</u> (p_1-p_2) =

[(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }, (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }]

= [ (0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }, (0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} } ]

 = [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0
3 years ago
How to solve this problem <br><br>.         0.3 [1.57 – (0.6)²]
love history [14]
Well (1.57- 0.6 X 0.6) x 0.3 = 0.363
3 0
3 years ago
Read 2 more answers
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