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Burka [1]
2 years ago
14

Please help thanks I appreciate it! :)

Mathematics
1 answer:
salantis [7]2 years ago
5 0

Answer:

Step-by-step explanation:

120 x .6 =72

120-72 = 48

48x .6 = 28.8

72+28.8 = 100.8

She sold approximately 101 cups of coffee

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Hii guys I am riyaz aly​
lesya [120]
Hello im Esmerallda
How are u
6 0
3 years ago
How do u set up the word problem
galina1969 [7]

Answer:

8 feet

Step-by-step explanation:

First, draw a picture.  The pole is 22 feet tall.  Katherine is 5.5 feet tall.  She is 24 feet from the pole.  The light from the top of the pole to the tip of her shadow forms two similar right triangles.  The length of her shadow is x.

Now use similar triangles to write a proportion:

22 / (24 + x) = 5.5 / x

Cross multiply:

22x = 5.5 (24 + x)

Distribute:

22x = 132 + 5.5x

Combine like terms:

16.5x = 132

Divide:

x = 8

The shadow is 8 feet long.

6 0
3 years ago
Solve the equation. Round to the nearest hundredth. Show work.
muminat

Answer:

Final answer is approx x=0.16.

Step-by-step explanation:

Given equation is 2.8\times 13^{4x} +4.8 = 19.3

Now we need to solve equation 2.8\times 13^{4x} +4.8 = 19.3 and round to the nearest hundredth.

2.8\times 13^{4x} +4.8 = 19.3

2.8\times 13^{4x} = 19.3-4.8

2.8\times 13^{4x} = 14.5

13^{4x} = \frac{14.5}{2.8}

13^{4x} = 5.17857142857

\log(13^{4x}) = \log(5.17857142857)

4x \log(13) = \log(5.17857142857)

4x = \frac{\log(5.17857142857)}{\log\left(13\right)}

4x = 0.641154659628

x = \frac{0.641154659628}{4}

x = 0.160288664907

Round to the nearest hundredth.

Hence final answer is approx x=0.16.

4 0
3 years ago
the marks in an examination for a Physics paper have normal distribution with mean μ and variance σ2 . 10% of the students obtai
Blababa [14]

Let X be the random variable for the number of marks a given student receives on the exam.

10% of students obtain more than 75 marks, so

P(X>75)=P\left(\dfrac{X-\mu}\sigma>\dfrac{75-\mu}\sigma\right)=P(Z>z_1)=0.10

where Z follows a standard normal distribution. The critical value for an upper-tail probability of 10% is

P(Z>z_1)=1-F_Z(z_1)=0.10\implies z_1=F_Z^{-1}(0.90)

where F_Z(z)=P(Z\le z) denotes the CDF of Z, and F_Z^{-1} denotes the inverse CDF. We have

z_1=F_Z^{-1}(0.90)\approx1.2816

Similarly, because 20% of students obtain less than 40 marks, we have

P(X

so that

P(Z

Then \mu,\sigma are such that

\dfrac{75-\mu}\sigma\approx1.2816\implies75\approx\mu+1.2816\sigma

\dfrac{40-\mu}\sigma\approx-0.8416\implies40\approx\mu-0.8416\sigma

and we find

\mu\approx53.8739,\sigma\approx16.4848

3 0
3 years ago
Help me and there will be an apple at your doorstep. (There might be some shipping issues)
Alexeev081 [22]

Answer:

9.30 maybe

Step-by-step explanation:

I just did mathz

3 0
3 years ago
Read 2 more answers
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