Please help me I need these answers

shirley has $540 un her bank account. She withdraws $35 each week to cover her expenses. write an equation that relates the amou

Veseljchak [2.6K]

540-35x= y

540 is the amount she has minus 35 and x is the weeks she withdrawn the money and y is the answer.

6 0

2 years ago

Find the distance between the points (-1,-5) and (-10,7).

Paladinen [302]

Answer:

15 units

Step-by-step explanation:

Calculate the distance d using the distance formula

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = (- 1, - 5) and (x₂, y₂ ) = (- 10, 7)

d = $\sqrt{(-10+1)^2+(7+5)^2}$

= $\sqrt{(-9)^2+12^2}$

= $\sqrt{81+144}$

= $\sqrt{225}$

= 15 units

5 0

2 years ago

Define z_alpha to be a z-score with an area of alpha to the right. For Example: z_0.10 means P(Z > z_0.10) = 0.10. We would a

Reptile [31]

Answer:

a) P(-z_0.025 < Z < z_0.025)

For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(0.025,0,1)"

And for this case the two values are : $z_{crit}= \pm 1.96$

b) P(-z_{\alpha/2} < Z < z_{\alpha/2})

For this case we want a quantile that accumulates $\alpha/2$ of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(alpha/2,0,1)"

c) For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got $z_{\alpha/2}=1.64$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Part a

P(-z_0.025 < Z < z_0.025)

For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(0.025,0,1)"

And for this case the two values are : $z_{crit}= \pm 1.96$

Part b

P(-z_{\alpha/2} < Z < z_{\alpha/2})

For this case we want a quantile that accumulates $\alpha/2$ of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(alpha/2,0,1)"

Part c

For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got $z_{\alpha/2}=1.64$

6 0

3 years ago

The daily average temperature in Santiago, Chile, varies over time in a periodic way that can be modeled approximately by a trig

natali 33 [55]

Answer:

a) the trigonometric function is;

$y = 7.5 sin ( \frac{2 \pi}{365}t + \frac{337 \pi}{730})+ 21.5$

b) $y = 28.36^0 \ C$ ( to two decimal places)

Step-by-step explanation:

This data can be represented by the sinusoidal function of the form :

$\mathbf{y = A sin (Bt -C)+D}$

where A = amplitude and which can be determined via the formula:

$A = \dfrac{largest \ temperature - lowest \ temperature}{2}$

$A = \dfrac{29-14}{2}$

$A = \dfrac{15}{2}$

A = 7.5° C

where B = the frequency;

Since the data covers a period of 3 days ; then $\dfrac{2 \pi}{B } =365$

$B = \dfrac{2 \pi}{365}$ ( where 365 is the time period )

The vertical shift is found by the equation D;

D = $\frac{largest \ temperature + lowest \ temperature}{2}$

D = $\frac{29+14}{2}$

D = 21.5

Replacing the values of A ; B and D into the above sinusoidal function; we have :

$y = 7.5 sin (\frac{2 \pi}{365}t -C) + 21.5$

From the question; when it is 7th of the year ( i.e January 7);

t = 7 and the temperature (y) = 29° C

replacing that too into the above equation; we have:

$29= 7.5 sin (\frac{2 \pi}{365}*7 -C) + 21.5$

$29= 7.5 sin (\frac{14 \pi}{365} -C) + 21.5$

$\frac{29-21.5}{7.5}= sin (\frac{14 \pi}{365} -C)$

$1= sin (\frac{14 \pi}{365} -C)$

$sin^{-1}(1)= (\frac{14 \pi}{365} -C)$

$\frac{\pi}{2}= (\frac{14 \pi}{365} -C)$

$C= (\frac{14 \pi}{365} -\frac{\pi}{2})$

$C= (\frac{28 \pi- 365 \pi}{730} )$

$C= \frac{-337 \pi}{730}$

Thus; the trigonometric function is;

$y = 7.5 sin ( \frac{2 \pi}{365}t + \frac{337 \pi}{730})+ 21.5$

Similarly; to determine the temperature o Jan 31; i.e when t= 31 ; we have :

$y = 7.5 sin ( \frac{2 \pi}{365}*31+ \frac{337 \pi}{730})+ 21.5$

$y = 7.5 sin ( \frac{62 \pi}{365}+ \frac{337 \pi}{730})+ 21.5$

$y = 7.5 sin ( \frac{124 \pi+ 337 \pi }{730})+ 21.5$

$y = 7.5 sin ( \frac{461 \pi }{730})+ 21.5$

$y = 7.5 *( 0.915)+ 21.5$

$y = 6.8689+ 21.5$

$y = 28.36^0 \ C$ ( to two decimal places)

7 0

2 years ago

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The after school craft center has 15 boxes of 64 crayons each. In 12 of the boxes , 28 of the crayons have not been used. All th

Korolek [52]

The <em><u>correct answer</u></em> is:

624 crayons.

Explanation:

In 12 of the boxes, 28 crayons have not been used; this leaves 64-28=36 crayons that have been used. 12(36) = 432 crayons have been used in these boxes.

3 full boxes have been used; this is 3(64) = 192 crayons.

Together this makes 432+192 = 624 crayons that have been used.

5 0

2 years ago

Read 2 more answers