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8_murik_8 [283]

2 years ago

10

Input Output Which ordered pair needs to be removed in order for the mapping to represent a function? O (-3, -4) O (-2, -1) O (1

, -3) O (3,7)

1 answer:

QveST [7]2 years ago

5 0

Answer:

(-2,-1)

Step-by-step explanation:

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What is the difference in place value between 2 in 324 and the 2 in 0.324

icang [17]

Answer: the decimal should still be there but you don’t want to have a zero so 3.42 would be correcr

Step-by-step explanation:

4 0

3 years ago

Help plz!!!!!!!!!!!!!!!!!!!!

RUDIKE [14]

4 Triangles 1/2(6)6()= 18 18x4= 72
5 squares 3X3=9. 5x9= 45

72+45= 117 sq in

5 0

3 years ago

A rectangular area adjacent to a river is fenced in; no fence is needed on the river side. The enclosed area is 1500 square fee

ZanzabumX [31]

Answer:

a) C(x) = 15000/x + 6x +80

b) Domain of C(x) { R x>0 }

Step-by-step explanation:

We have:

Enclosed area = 1500 ft² = x*y from which y = 1500 / x (a) where x is perpendicular to the river

Cost = cost of sides of fenced area perpendicular to the river + cost of side parallel to river + cost of 4 post then

Cost = 10*y + 2*3*x + 4*20 and accoding to (a) y = 1500/x

Then

C(x) = 10* ( 1500/x ) + 6*x + 80

C(x) = 15000/x + 6x +80

Domain of C(x) { R x>0 }

5 0

2 years ago

What is the area of a flour tortilla that has a 14-centimeter radius?

kykrilka [37]

The formula for the area of a circle is
$\pi {r}^{2}$
so you simply plug your value for radius into the equation!
$\pi {(14)}^{2}$
Square 14
$196\pi$
And Multiply by pi!
615.75 cm^2 approximately!
I hope this helps!

5 0

2 years ago

How many times must we toss a coin to ensure that a 0.95-confidence interval for the probability of heads on a single toss has l

musickatia [10]

Answer:

(1) 97

(2) 385

(3) 9604

Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for population proportion is:

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

The margin of error in this interval is:

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

The formula to compute the sample size is:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}$

(1)

Given:

$\hat p = 0.50\\MOE=0.1\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.1^{2}}\\=96.04\\\approx97$

Thus, the minimum sample size required is 97.

(2)

Given:

$\hat p = 0.50\\MOE=0.05\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.05^{2}}\\=384.16\\\approx385$

Thus, the minimum sample size required is 385.

(3)

Given:

$\hat p = 0.50\\MOE=0.01\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.01^{2}}\\=9604$

Thus, the minimum sample size required is 9604.

8 0

2 years ago