All categories

3 years ago

Is 1/3 less than 3/4 and is 9/16 greater than 3/4

Mathematics

1 answer:

Mandarinka [93]3 years ago

8 0

1/3 is less than 3/4 but i am pretty sure 9:16 is less than 3/4 because 9/16 is close to 8/16 which is 1/2

You might be interested in

Someone help meh plzz

madam [21]

Answer:

1) 229 Teens

2) (12,-2)

6 0

3 years ago

I will appreciate if you answer this problem,thanks

aliya0001 [1]

Answer:

None

Step-by-step explanation:

By graphing out the both line segments through graphing software, segments AB and CD have no intersections on the co-ordinate plane.

I've attached a screenshot of what i graphed out.

3 0

3 years ago

For each level of precision, find the required sample size to estimate the mean starting salary for a new CPA with 95 percent co

Rzqust [24]

Answer:

(a) Margin of error ( E) = $2,000 , n = 54

(b) Margin of error ( E) = $1,000 , n = 216

(c) Margin of error ( E) = $500 , n= 864

Step-by-step explanation:

Given -

Standard deviation $\sigma$ = $7,500

$\alpha$ = 1 - confidence interval = 1 - .95 = .05

$Z_{\frac{\alpha}{2}}$ = $Z_{\frac{.05}{2}}$ = 1.96

let sample size is n

(a) Margin of error ( E) = $2,000

Margin of error ( E) = $Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

E = $Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}$

Squaring both side

$E^{2} = 1.96^{2}\times\frac{7500^{2}}{n}$

$n =\frac{1.96^{2}}{2000^{2}} \times 7500^{2}$

n = 54.0225

n = 54 ( approximately)

(b) Margin of error ( E) = $1,000

E = $Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

1000 = $Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}$

Squaring both side

$1000^{2} = 1.96^{2}\times\frac{7500^{2}}{n}$

$n =\frac{1.96^{2}}{1000^{2}} \times 7500^{2}$

n = 216

E = $Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

500 = $Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}$

Squaring both side

$500^{2} = 1.96^{2}\times\frac{7500^{2}}{n}$

$n =\frac{1.96^{2}}{500^{2}} \times 7500^{2}$

n = 864

7 0

3 years ago

DOES ANYONE UNDERSTAND QUADRATIC APPLICATIONS FOR ALGEBRA 1???? PLS DONT RESOND IF YOU DINT KNOW

lilavasa [31]

Answer:

1 ) Vertex form.

2 ) Standard form.

3 ) I'm not sure about this one, but there are only 3 forms, in which the last one is factored form, and the other two are above, so it must be factored form.

Step-by-step explanation:

1 ) Vertex form : y = a( x - h )² + k

Vertex form shows you where the vertex(the lowest/highest point aka ( h, k ) ) of the parabola is, without graphing it. (But you could graph it and find the vertex like that, I recommend you use desmos graphing)

2 )Standard form : y = ax² + bx + c

Standard form can be used to show the y-intercept( aka c ), in this case, where the snowball was thrown from. (Again, you can graph it and find where he parabola hits the y axis)

3 ) Factored form : ( ax + c )( bx + d )

Factored form tells us the x-intercepts, or in this case where the snowball landed or how long the snowball is in the air. (like i said, you could graph it, and find the points(there could be only one or maybe none in some cases) that are on the x axis)

I know this was four days ago, but I hope this helped you understand a little better.

4 0

2 years ago

Can I have help solving this?

tensa zangetsu [6.8K]

4^2 + 2(4 + 6)

4^2 + 2(10)

16 + 2(10)

16 + 20

3 0

3 years ago