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3 years ago

9

Lincoln's monthly bank statement showed the following deposits and withdrawals:

1 answer:

gayaneshka [121]3 years ago

8 0

First you need to add the yellow pingien the ask the red turtule for the green panda name ok

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What is -3.7 + 2.8 <br> equal too

Ber [7]

Answer:

-0.9

Step-by-step explanation

Trust <3

7 0

3 years ago

Refer to the random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.50. a.

zysi [14]

Answer:

a) $78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

b) $78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

d) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

Step-by-step explanation:

Part a

For this case we have the following data given

$\bar X = 78.25$ represent the sample mean for the customer order totals

$\sigma =22.50$ represent the population deviation

$n= 40$ represent the sample size selected

The confidence level is 90% or 0.90 and the significance level would be $\alpha=0.1$ and $\alpha/2 = 0.05$ and the critical value from the normal standard distirbution would be given by:

$z_{\alpha/2}=1.64$

And the confidence interval is given by:

$\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

Part b

The sample size is now n = 75, but the same confidence so the new interval would be:

$78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

Part c

For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

Part d

The margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

3 0

3 years ago

What the slope-intercept form?<br><br> Slope, m = 2 <br> y-intercept, b = 5

Lorico [155]

Answer:

y = mx + b

so its

y = 2x + 5

8 0

3 years ago

What is the length of the hypotenuse? If necessary, round to the nearest tenth.

Hitman42 [59]

<u>Answer:</u>

The length of the hypotenuse is 40 mm.

<u>Step-by-step explanation:</u>

<em>We can solve for 'c' using Pythagoras theorem. Let's solve it.</em>

c² = 32² + 24²
=> c² = 1024 + 576
=> c² = 1600
=> c = 40

Hence, the length of the hypotenuse is 40 mm.

Hoped this helped.

$BrainiacUser1357$

7 0

3 years ago

From the sum of a2−2ab+b2 and 2a2+2ab+b2 subtract the sum of a2−b2 and a2+ab+3b2

Sedaia [141]

Answer:

$a^2-ab$

Step-by-step explanation:

We need to find the sum of $a^2-2ab+b^2$ and $2a^2+2ab+b^2$ first.

Adding $a^2-2ab+b^2$ + $2a^2+2ab+b^2$

Combining like terms, we get

$a^2+2a^2=3a^2$

-2ab+2ab = 0

$b^2+b^2=2b^2$

Therefore,

$a^2-2ab+b^2 +2a^2+2ab+b^2=3a^2+2b^2$ .

Now, we need to find the sum of $a^2-b^2 \ \ and \ \ a^2+ab+3b^2.$ .

Adding $a^2−b^2+a^2+ab+3b^2$ .

Combining like terms, we get

$a^2+a^2=2a^2$

$-b^2+3b^2=2b^2$ .

Therefore,

$a^2−b^2+a^2+ab+3b^2=2a^2+ab+2b^2.$

Now, subtracting

$3a^2+2b^2 -(2a^2+ab+2b^2)$ .

Distributing minus sign over second parenthesis, we get

$3a^2+2b^2-2a^2-ab-2b^2$ .

Combining like terms,

$3a^2-2a^2=a^2$

$2b^2-2b^2=0$

Therefore,

$3a^2+2b^2-2a^2-ab-2b^2=a^2-ab$ .

Therefore, the difference of the sum of $a^2-2ab+b^2$ + $2a^2+2ab+b^2$ and $a^2-b^2+a^2+ab+3b^2.$ is $a^2-ab.$

5 0

3 years ago