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3 years ago

PLEASE HELP WITH THESE ❤️ (DUE TODAY!!!)

Mathematics

1 answer:

umka21 [38]3 years ago

8 0

Answer

Caythlyn patter will be

mina pattern will be

Step-by-step explanation:

all your doing is multiply and then adding because an number time 1 is the number it is multiplied by. and then keep adding and mutiply on to find you answer

(p.s im not sure so i hope its right I tried)

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4x(2x -3) -5(3x -4) simplifyed

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Answer:

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3 years ago

Please answer this question !! BRAINLIEST !!

Misha Larkins [42]

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4 years ago

3 ÷ 3/4 (3/4 Is a fraction

kogti [31]

Answer:

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3 years ago

Read 2 more answers

What is (-33x-78)+(5x8x9x10)divided by9.75

kherson [118]

Answer:

6,174

Step-by-step explanation:

(-33 * - 78) + (5 * 8 * 9 * 10)

Remove the brackets:

-33 * - 78 + 5 * 8 * 9 * 10

Solve like so:

-33 * - 78 = 2,574

5 * 8 * 9 * 10 = 3,600

(2,574) + (3,600)

2,574 + 3,600 = 6,174

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4 years ago

Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 1

Zarrin [17]

Answer:

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean age of selected students = 22.4 years

s = sample standard deviation = 3.8 years

n = sample of students = 19

$\mu$ = population mean

Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.878 < $t_1_8$ < 2.878) = 0.99 {As the critical value of t at 18 degree of

freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.878) = 0.99

P( $-2.878 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.878 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ]

= [ $22.4 -2.878 \times {\frac{3.8}{\sqrt{19} }$ , $22.4 +2.878 \times {\frac{3.8}{\sqrt{19} }$ ]

= [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

6 0

3 years ago