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2 years ago

Help me please... I have no clue on how to do this..

Mathematics

2 answers:

diamong [38]2 years ago

7 0

The answer is B.

To turn a decimal into a percent, just divide it by 100 and add a % sign.

Hope this helped! ♡

tresset_1 [31]2 years ago

3 0

D is the answer because in the text it says 20.95% and that is the same as D

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4(g - 1 )=24 what is the answer

babunello [35]

Hello There!

First Expand:
4g - 4 = 24
Then solve:
4g = 24 + 4
4g = 28
g = 28/4
g = 7.

Hope This Helps You!
Good Luck :)

6 0

2 years ago

Read 2 more answers

PLEASE HELP ME!

dezoksy [38]

Thicknesses at different point are: 41, 38, 36, 29, 34, 44, 46, 43, 35, 40

In increasing order: 29, 34, 35, 36, 38, 40, 41, 43, 44, 46

Median = (38+40)/2 = 39m

Median thickness is 39m

5 0

2 years ago

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19x+rx=-37+w solve for x

ipn [44]

Factor the left side:-

x (19 + r) = -37 + w

x = ( -37 + w) / (19 + r) Answer

3 0

2 years ago

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Roberto wants to buy a $40 gift for his father. He finished one odd job in 4 hours at $3.75 an hour. He is now working at a job

Nikitich [7]

Step-by-step explanation:

amount left for the money to reach $40

=40-3.75

=36.25

using a method of proportion,

If $4 : 1 hour,then

$36.25 : ?

if more, less divides.

= 36.25 / 4 ×1

=36.25/4

=9.0625 approximately 9 hours

4 0

2 years ago

Suppose the time a child spends waiting at for the bus as a school bus stop is exponentially distributed with mean 7 minutes. De

Gala2k [10]

Answer:

The probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

Step-by-step explanation:

Let the random variable X represent the time a child spends waiting at for the bus as a school bus stop.

The random variable X is exponentially distributed with mean 7 minutes.

Then the parameter of the distribution is, $\lambda=\frac{1}{\mu}=\frac{1}{7}$ .

The probability density function of X is:

$f_{X}(x)=\lambda\cdot e^{-\lambda x};\ x>0,\ \lambda>0$

Compute the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning as follows:

$P(6\leq X\leq 9)=\int\limits^{9}_{6} {\lambda\cdot e^{-\lambda x}} \, dx$

$=\int\limits^{9}_{6} {\frac{1}{7}\cdot e^{-\frac{1}{7} \cdot x}} \, dx \\\\=\frac{1}{7}\cdot \int\limits^{9}_{6} {e^{-\frac{1}{7} \cdot x}} \, dx \\\\=[-e^{-\frac{1}{7} \cdot x}]^{9}_{6}\\\\=e^{-\frac{1}{7} \cdot 6}-e^{-\frac{1}{7} \cdot 9}\\\\=0.424373-0.276453\\\\=0.14792\\\\\approx 0.148$

Thus, the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

6 0

2 years ago