All categories

3 years ago

I need help please on this math problem

Mathematics

2 answers:

My name is Ann [436]3 years ago

4 0

Answer: 72

Step-by-step explanation:Do 6x3x4.

yawa3891 [41]3 years ago

3 0

L=6 cubes
w=3 cubes
h=4 cubes
v=lwh
= (6)(3)(4)
= 72 cubes

You might be interested in

Help me fellow idiots (jk)

Ierofanga [76]

Answer:

$\frac{4}{5}$

8 0

3 years ago

Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in

Leno4ka [110]

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - α)% confidence interval for the population variance is given as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

It is provided that:

n = 20

s = 3.9

Confidence level = 95%

⇒ α = 0.05

Compute the critical values of Chi-square:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907$

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

$\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45$

Thus, the 95% confidence interval for the population variance is (8.80, 32.45).

4 0

3 years ago

Justin has $7.50 more than Eva, and Emma has $12 less than Justin. Together, they have a total of $63.00. How much money does ea

uysha [10]

Answer:

Step-by-step explanation: emma's amount be X . justin has $7.50+x because he has $7.50 more than emma.Since emma has 7.50+x-$12.to get everyones amount add what they have together which is=$7.50+x+$7.50+x-12 which in total is $63.00.after the calculations you get x=30.now distribute the x equally. so justin has $7.50+30=37.50.we also distribute it with emma's which is 7.50+30-12=25.5.

justin=$37.50

Emma=$25.50

to be sure add 37.50+25.50=$63

5 0

4 years ago

Quadrilateral A'B'C'D' is the image of quadrilateral ABCD under a rotation about the origin, (0,0). y 71 c 3+ 2+ 2 1 2 3 4 5 6 7

VLD [36.1K]

Answer:

give me the bubble bubble I really do not know

7 0

2 years ago

Find the area of each regular polygon. Round your answer to the nearest tenth if necessary.

tatuchka [14]

*I am assuming that the hexagons in all questions are regular and the triangle in (24) is equilateral*

(21)

Area of a Regular Hexagon: $\frac{3\sqrt{3}}{2}(side)^{2} = \frac{3\sqrt{3}}{2}*(\frac{20\sqrt{3} }{3} )^{2} =200\sqrt{3}$ square units

(22)

Similar to (21)

Area = $216\sqrt{3}$ square units

(23)

For this case, we will have to consider the relation between the side and inradius of the hexagon. Since, a hexagon is basically a combination of six equilateral triangles, the inradius of the hexagon is basically the altitude of one of the six equilateral triangles. The relation between altitude of an equilateral triangle and its side is given by:

$altitude=\frac{\sqrt{3}}{2}*side$