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3 years ago

Find the 60th term of the arithmetic sequence 4,-1,-6-

Mathematics

1 answer:

kobusy [5.1K]3 years ago

8 0

Answer:

-291

Step-by-step explanation:

The common difference is -5

So the 60th term of the arithmetic sequence is

4+(60-1)×-5

=4+59×-5

=4+-295

=-291

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Neporo4naja [7]

A= l×w 4700=94×? 4700÷94=50 ~jz Hope it helps

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3 years ago

What is 25 divided by its square root? Brainliest for best answer.

uysha [10]

Answer:

Step-by-step explanation:

Because the square root of 25 is 5 so then do 25 divided by 5 to get 5 as your awnser.

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4 years ago

Many smartphones, especially those of the LTE-enabled persuasion, have earned a bad rap for exceptionally bad battery life. Batt

BartSMP [9]

Answer:

Null hypothesis: the variance in hours of usage for talking is not greater than the the variance in hours of usage for internet.

$H_o$ : $\sigma _1 ^2 \leq \sigma _2^2$

Alternative hypothesis: the variance in hours of usage for talking is greater than the the variance in hours of usage for internet.

$H_a : \sigma_1^2 > \sigma_2^2$

$\mathbf{s_ 1 =16.11}$

$\mathbf{s_2 = 7.98}$

Step-by-step explanation:

Let $x_1$ and $x_2$ be the two variables that represents the battery life in hours for talking usage and battery life in hours for internet usage respectively.

The hypothesis can be formulated as:

Null hypothesis: the variance in hours of usage for talking is not greater than the the variance in hours of usage for internet.

$H_o$ : $\sigma _1 ^2 \leq \sigma _2^2$

Alternative hypothesis: the variance in hours of usage for talking is greater than the the variance in hours of usage for internet.

$H_a : \sigma_1^2 > \sigma_2^2$

The standard deviation for the battery usage for talking is :

$\bar x_1 = \dfrac{1}{n_1} \sum x_i \\ \\ \bar x_1 = \dfrac{1}{12}(35.8 +22.4+...+35.5) \\ \\ \bar x_1 = \dfrac{241.2}{12} \\ \\ \bar x_1 =20.1$

The standard deviation Is:

$s_ 1 = \sqrt{\dfrac{1}{n_1-1}\sum (x{_1i}-\bar x_i)^2}$

$s_ 1 = \sqrt{\dfrac{1}{12-1}\sum (35.8- 20.1)^2+ (35.5-20.1)^2}$

$s_ 1 = \sqrt{259.568}$

$\mathbf{s_ 1 =16.11}$

The standard deviation for the battery life usage for the internet is :

$\bar x_2 = \dfrac{1}{n_2} \sum x_{2i}$

$\bar x_2 = \dfrac{1}{10} (24.0+12.5+36.4+...+4.7})$

$\bar x_2 = \dfrac{115}{10}$

$\bar x_2 = 11.5$

Thus; the standard deviation is:

$s_2 = \sqrt{\dfrac{1}{n_2-1}(x_{2i}- \bar x_2)^2}$

$s_2 = \sqrt{\dfrac{1}{10-1}(24-11.5)^2+(4.7-11.5)^2}$

$s_2 = \sqrt{63.60}$

$\mathbf{s_2 = 7.98}$

4 0

3 years ago

Select the correct answer. Consider functions f and g.

vlabodo [156]

C) 1 b =-2. D) 7 f(-2) = 3(-8) - 2. -5 x=6
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2 years ago