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valentina_108 [34]
3 years ago
13

What is the density of a brownie the shape of a cube weighing 15 grams measuring 5 cm on a side?

Mathematics
1 answer:
ipn [44]3 years ago
5 0

Answer:

  0.12 g/cm³

Step-by-step explanation:

Density is the ratio of mass to volume. The volume of the brownie is the cube of its side dimension:

  V = s³ = (5 cm)³ = 125 cm³

Then the density is ...

  ρ = M/V = (15 g)/(125 cm³) = 0.12 g/cm³

The density of the brownie is 0.12 g/cm³.

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Random samples of size 64 are taken from an infinite population whose mean and standard deviation are 24 and 6.4, respectively.
umka2103 [35]

Answer:

Mean 24, standard error 0.8

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation, also called standard error s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 24, \sigma = 6.4, n = 64

What are the mean and the standard error of the sample mean?

By the Central Limit Theorem, mean 24 and standard error s = \frac{6.4}{\sqrt{64}} = 0.8

7 0
3 years ago
4-b/12=-3<br> What's b equal
NNADVOKAT [17]

Answer:

19-12=7, so 4-7=-3, so b is equal to 19

7 0
3 years ago
An item has a listed price of $65 . If the sales tax rate is 3%, how much is the sales tax (in dollars)?
larisa86 [58]

Answer:

$1.95         hope this helped

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Find the value of x (pic attached)
nlexa [21]
Answer: x=4, I added a photo of my work in case you need it I hope it helps

6 0
4 years ago
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Find the area of triangle ABC with vertices A(2,1), B (12,2), C (12,8). Hence, or otherwise find the perpendicular distance from
Lapatulllka [165]
<h2>The area of a triangle is =54 square units</h2><h2>The perpendicular distance from B to AC is = \frac{108}{\sqrt{149} } units</h2>

Step-by-step explanation:

Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)

x_1=2,y_1=1,x_2=12,y_2=2,x_3=12 and      y_3=8

The area of a triangle is= \frac{1}{2} [x_1(y_2-y_3) +x_2 (y_3- y_1)+x_3(y_1-y_2)]

=|\frac{1}{2} [2(2-8+12(8-1)+12(1-2)]|

=|-54| = 54 square units

The length of AC = \sqrt{(x_1-x_3)^{2} +(y_1-y_3)^2}

                          = \sqrt{(2-12)^{2} +(1-8)^2}

                         =\sqrt{149} units

Let the perpendicular distance from B to AC be = x

According To Problem

\frac{1}{2} \times  x \times \sqrt{149} = 54

⇔x =\frac{108}{\sqrt{149} } units

Therefore the perpendicular distance from B to AC is = \frac{108}{\sqrt{149} } units

5 0
3 years ago
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