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3 years ago

Find the distance between the points (5, -2) and (1, -4)

Mathematics

2 answers:

Gemiola [76]3 years ago

3 0

Answer:

√20

Step-by-step explanation:

d = √(x₂-x₁)²+(y₂-y₁)²

d = √(1-5)²+(-4+2)²

d = √(-4)²+(-2)²

d = √16+4

d = √20

Natasha_Volkova [10]3 years ago

3 0

Answer:

4.472 or √20

Step-by-step explanation:

Use the distance formula and plug in your x and y points, and then you should have your answer. Distance formula: d= √(x2+x1)^2 + (y2+y1)^2

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marusya05 [52]

Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

$n= 955$ represent the sampel size slected

$x = 812$ number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

$\hat p =\frac{955-812}{955}= 0.150$

The confidence interval for the proportion would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the significance is $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution and we got.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131$

$0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169$

And the 90% confidence interval would be given (0.131;0.169).

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