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2 years ago

Can someone please help me with this :(

Mathematics

1 answer:

mrs_skeptik [129]2 years ago

5 0

Answer:

first find the surface area of one of the square and one of the triangle.

then multiple it by the number of square and triangle they have

finally add the total surface area of the triangle with the total surface area of the square.

Step-by-step explanation:

for one square

area= base * width

= 10*10

=100

total area of square= number of square * area of one square

t=6*100

t=600

for one triangle

area = 1/2 base * height

= 1/2 *10*8.66

=43.3

total area of triangle=number of triangle* area of one triangle

t2=8*43.3

=346.4

total surface area= total surface area of the square+ total surface area of the triangle

=600+346.4

=946.4 in^2

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9.4 The heights of a random sample of 50 college stu- dents showed a mean of 174.5 centimeters and a stan- dard deviation of 6.9

gladu [14]

Answer: a) (176.76,172.24), b) 0.976.

Step-by-step explanation:

Since we have given that

Mean height = 174.5 cm

Standard deviation = 6.9 cm

n = 50

we need to find the 98% confidence interval.

So, z = 2.326

(a) Construct a 98% confidence interval for the mean height of all college students.

$x\pm z\times \dfrac{\sigma}{\sqrt{n}}\\\\=(174.5\pm 2.326\times \dfrac{6.9}{\sqrt{50}})\\\\=(174.5+2.26,174.5-2.26)\\\\=(176.76,172.24)$

(b) What can we assert with 98% confidence about the possible size of our error if we estimate the mean height of all college students to be 174.5 centime- ters?

Error would be

$\dfrac{\sigma}{\sqrt{n}}\\\\=\dfrac{6.9}{\sqrt{50}}\\\\=0.976$

Hence, a) (176.76,172.24), b) 0.976.

8 0

2 years ago

Without calculating, which has a bigger volume. A cube that has a length, width, and height of 18 m. Or a sphere with a radius o

Evgesh-ka [11]

Weird. A period appears above this... huh.

Answer:

[Th]e cube has a greater value.

Step-by-step explanation:

What the word problem really wants us to get [is ]the question of 'Which is greater, A=6a^2 when 'a' [is] 18 or A=4 $\pi$ r^2 when r = 9? And here's how to solve that.

Starting with the[ c]ube we have A=6a^2. A bit t[o]o simple, right?

A=6(18)^2 Substitute numbers.

A=6(324) Solve ex[p]onents.

A=1944 Mult[i]ply.

So w[e] know that the cube is 1944 meters cube[d ] in area. But what about the more [f]ormidable sphere? Fo[r] it we need a slightly m[o]re co[m]plicated formula, A=4 $\pi$ r^2. However, instead of using the real pi I will be rounding to 3.14, since we have no calculator so anything more would take way too long and fry your[ bra]in.

A=4(3.14)(9)^2 Subst[i]tute numbers.

A=4(3.14)(81) Solve expone[n]ts.

A=12.56(81) Multip[ly].

A=1017.36 Multiply again[.]

Now, since I'm sure all of us can count, we know that 1944 is greater than 1017.36. Or in other words, the cube is bigger than the sphere.

And PLEASE don't copy this guys. Make your own iteration. Change it up!

3 0

2 years ago

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Answer:

Step-by-step explanation: