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3 years ago

Abhivardhansinghrathore7979

Mathematics

1 answer:

mario62 [17]3 years ago

8 0

Answer:

hlo

Step-by-step explanation:

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Are the ratios 16:9 and 2:1 equivalent?<br> yes<br> no

yarga [219]

Answer:

Step-by-step explanation:

To be equivalent it would have to be 16/8 not 16/9

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3 years ago

The answer is 3/2x but I don’t know how to solve it... please explain

lesya [120]

Answer:

y = 3/2x by making use of angle relationships in triangles

Step-by-step explanation:

Here's one way to solve it.

∠ADE is an external angle to ΔBDE. As such, its measure will be the sum of the measures of the remote interior angles, ∠DBE and ∠DEB:

∠ADE = 2x° +y°

If we call the intersection point of AC and DE point G, then ∠AGE is an exterior angle to ΔADG. As such, its measure is the sum of the remote interior angles:

∠AGE = ∠GAD +∠GDA

3y° = x° +(2x° +y°)

2y = 3x . . . . . . . . . . subtract y°, collect terms, divide by °

y = (3/2)x . . . . . . . . divide by 2

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3 years ago

Which graph represents the solution set of the system of inequalities? {y≤2x+1y>−2x−3

Svet_ta [14]

The figure shown below graphs the inequalities
y ≤ 2x + 1
y > -2x - 3

The shaded area shows the solution region.

7 0

4 years ago

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docker41 [41]

Answer

yes it is correct

Step-by-step explanation:

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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

Abhivardhansinghrathore7979​

Abhivardhansinghrathore7979