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soldi70 [24.7K]

2 years ago

11

A study of 860 U.S. ATMs (automated teller machines) found that the average surcharge for withdrawals from a competing bank was

$1.15. (Source: Public Interest Research Groups) The sample is ATMs from a competing bank. Question 14 options: True False

2 answers:

dexar [7]2 years ago

6 0

Im pretty sure your answer is true.

Valentin [98]2 years ago

6 0

Answer:

Yes this is true.

Step-by-step explanation:

Given is - A study of 860 U.S. ATMs found that the average surcharge for withdrawals from a competing bank was $1.15. The sample is ATMs from a competing bank.

A competing bank cannot be thousands. Rather a competing bank can have hundreds of ATM's throughout US. So, the given number of ATM's is sample here. The statement is true.

If it would have been that the sample ATM's is from all U.S, then the statement would have been false.

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A board 109 cm long has to be cut into two pieces so that the longer part is 10 cm more than twice as long as the shorter part.

11111nata11111 [884]

x = longer section
y = shorter section

x + y = 109
x = 2y + 10

so we sub in eq 2 into eq 1
2y + 10 + y = 109
3y + 10 = 109
3y = 109 - 10
3y = 99
y = 99/3
y = 33 <=== shorter section is 33 cm

x = 2y + 10
x = 2(33) + 10
x = 66 + 10
x = 76 <=== longer section is 76 cm

4 0

2 years ago

Given: △ABC, AB=5 square root 2 <br><br>m∠A=45°, m∠C=30°<br><br>Find: BC and AC

hichkok12 [17]

Answer:

Part a) $BC=10\ units$

Part b) $AC=13.66\ units$

Step-by-step explanation:

step 1

Find the length side BC

Applying the law of sines

we know that

$\frac{AB}{sin(C)}=\frac{BC}{sin(A)}$

we have

$AB=5\sqrt{2}\ units$

$A=45^o$

$C=30^o$

substitute

$\frac{5\sqrt{2}}{sin(30^o)}=\frac{BC}{sin(45^o)}$

solve for BC

$BC=\frac{5\sqrt{2}}{sin(30^o)}(sin(45^o))$

$BC=10\ units$

step 2

Find the measure of angle B

we know that

The sum of the interior angles in a triangle must be equal to 180 degrees

so

$m\angle A+m\angle B+m\angle C=180^o$

substitute the given values

$45^o+m\angle B+30^o=180^o$

$75^o+m\angle B=180^o$

$m\angle B=180^o-75^o$

$m\angle B=105^o$

step 3

Find the length side AC

Applying the law of sines

we know that

$\frac{AB}{sin(C)}=\frac{AC}{sin(B)}$

we have

$AB=5\sqrt{2}\ units$

$A=45^o$

$B=105^o$

substitute

$\frac{5\sqrt{2}}{sin(30^o)}=\frac{AC}{sin(105^o)}$

solve for AC

$AC=\frac{5\sqrt{2}}{sin(30^o)}(sin(105^o))$

$AC=13.66\ units$

4 0

2 years ago

37.5 x 10 to the negative 6 in scientific notation?

mestny [16]

The scientific notation is 37.5×10^-6

7 0

3 years ago

How many times larger is 5x 10^6than 5 x 10^2

svetoff [14.1K]

Answer:

To inf and beond

Step-by-step explanation:

6 0

2 years ago

Find a vector function that represents the curve of intersection of the paraboloid z=3x2+2y2 and the cylinder y=6x2. Use the var

Cerrena [4.2K]

Answer:

$\vec{r(t)}=t\hat{i}+6t^2\hat{j}+(3t^2+72t^4)\hat{k} \text{ For }t \in (-\infty, \infty)$

Step-by-step explanation:

First plug the equation of y into the equation of z, so that we get their intersection:

We plug

$y=6x^2$

Into: $z=3x^2+2y^2$

So, we get:

$z=3x^2+2(6x^2)^2\\z=3x^2+2(36x^4)\\z=3x^2+72x^4$

Then we set

$x=t$

And plug that in the equation $y=6x^2$ and the one we just got for z.

So that we get:

$y=6t^2, z=3t^2+72t^4$

Therefore, the vector function that represents the curve of intersection is:

$\vec{r(t)}=t\hat{i}+6t^2\hat{j}+(3t^2+72t^4)\hat{k} \text{ For }t \in (-\infty, \infty)$

3 0

2 years ago