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3 years ago

-30=-16+N what is he answer to this

Mathematics

1 answer:

tankabanditka [31]3 years ago

5 0

Answer:

N= -14

Step-by-step explanation:

first you add 16 to both sides, then -30 + -16 = -14

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Mademuasel [1]

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3 years ago

What is the nth term rule of the quadratic sequence below? 7, 14, 23, 34, 47, 62, 79

Julli [10]

Answer:

Tn = Tn-1 + 2(n-1) + 5

Kindly note that Tn-1 means T subscript n-1

Step-by-step explanation:

Here, we want an expression for the nth term.

First term is 7

Then first common difference is 7

second common difference is 7 + 2

Third common difference is 9 + 2

So within the common differences, the nth term is 7 + (n-1) 2

Now, the nth term of the series would be;

Tn = Tn-1 + 7 + 2(n-1)

Tn = Tn-1 + 7 + 2n -2

Tn = Tn-1 + 2n + 5

Now there is a fix to this,

n for the term is not the same n for the common difference.

the 7th term works with the 6th common. difference, while the 8th term work for the 7th common difference.

So we might need to rewrite the final expression as follows;

Tn = Tn-1 + 2(n-1) + 5

6 0

3 years ago

What’s the answer for 6?

horrorfan [7]

I'm not 100 percent sure but I think is e

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3 years ago

Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):

Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, $\mu_X$ represent the population mean for Brand B and let $\mu_Y$ represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

P.Q. = $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t_n__1+n_2-2$

where, $Xbar$ = Sample mean for Brand B data = 36.9

$Ybar$ = Sample mean for Brand A data = 32.9

$n_1$ = Sample size for Brand B data = 13

$n_2$ = Sample size for Brand A data = 9

$s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} }$ = 3.013

Here, $s^{2}_X$ and $s^{2} _Y$ are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < $t_2_0$ < 2.528) = 0.98

P(-2.528 < $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 2.528) = 0.98

P(-2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(Xbar -Ybar) -(\mu_X-\mu_Y)$ < 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

P( (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(\mu_X-\mu_Y)$ < (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98