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Lilit [14]
3 years ago
8

Evaluate 14y3 = 2 + 901 for y=6* Your answer​

Mathematics
1 answer:
svp [43]3 years ago
3 0

Answer:

hxhGmzgkzjzjzultstusutsyisyidyxyixyostosyosyisuoxuxhckcifidususu

Step-by-step explanation:

yeesyhdbdhdhdufhfhfbfbdbdvdvdvxbcbdhdhdhshssbbssgOysyizykxlhdloyzkhxkgxigsoyxly khxiyyykxkg

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Anyone smart at math?
RoseWind [281]

Answer:

The function f(x) = 4.95x represents the cost to purchase x pounds of chocolate from a candy company. The function g(x) = 0.99x + 2.25 models the cost to ship x pounds of chocolate from the company. Write a function, h(x), that models the total cost to purchase and ship x pounds of chocolate from the company.

Step-by-step explanation:

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8 0
3 years ago
Need help on angles of polygons please
aleksandrvk [35]

Answer:

x+87+92+105+135=540(sum of interior angle of pentagon)

x+419=540

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x=121

Step-by-step explanation:

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3 years ago
Given: BD = BF DE ⊥ BC , FK ⊥ AB Prove: ED ≅ FK
sergeinik [125]

This is the picture I am not very good at proofs but hopefully someone can solve it.

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3 years ago
Read 2 more answers
This two-word phrase indicates a ratio or fraction
svlad2 [7]
Ratio-Proportion is the answer
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4 years ago
Please I need help with this question!
likoan [24]

Answer:

I. 60%

II. 75.4 kg

Step-by-step explanation:

We will use the z-scores and the standard normal distribution to answer this questions.

We have a normal distribution with mean 69 kg and variance 25 kg^2 (therefore, standard deviation of 5 kg).

I. What percentage of adult male in Boston weigh more than 72 kg?

We calculate the z-score for 72 kg and then calculate the associated probability:

z=\dfrac{X-\mu}{\sigma}=\dfrac{72-69}{5}=\dfrac{3}{5}=0.6\\\\\\P(X>72)=P(z>0.6)=0.274

II.  What must an adult male weigh in order to be among the heaviest 10% of the population?

We have to calculate tha z-score that satisfies:

P(z>z^*)=0.1

This happens for z=1.28 (see attachment).

Then, we can calculate the weight using this transformation:

X=\mu+z^*\cdot\sigma=69+1.28\cdot 5=69+6.4=75.4

5 0
3 years ago
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