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krek1111 [17]
3 years ago
15

Which expressions are equivalent to the one below? Check all that apply.

Mathematics
1 answer:
Aleksandr [31]3 years ago
5 0

Answer:

D

Step-by-step explanation:

(5·5)*x = 25*x =25^x

the rest are wrong

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Solve equation: –6 + 3x = –9
ICE Princess25 [194]
3x=-9+6
3x=-3
x=-3/3
x=-1
3 0
3 years ago
Read 2 more answers
A poster is to have a total area of 125 cm2. There is a margin round the edges of 6 cm at the top and 4 cm at the sides and bott
jeka57 [31]

Answer:

The correct answer is 15 cm.

Step-by-step explanation:

Let the width of the required poster be a cm.

We need to have a 6 cm margin at the top and a 4 cm margin at the bottom. Thus total margin combining top and bottom is 10 cm.

Similarly total margin combining both the sides is (4+4=) 8 cm.

So the required printing area of the poster is given by {( a-10 ) × ( a - 8) } cm^{2}

This area is equal to 125 cm^{2} as per as the given problem.

∴ (a - 10) × (a - 8) = 125

⇒ a^{2} - 18 a +80 -125 =0

⇒ a^{2} - 18 a -45 = 0

⇒ (a-15) (a-3) = 0

By law of trichotomy the possible values of a are 15 and 3.

But a=3 is absurd as a > 4.

Thus the required answer is 15 cm.

7 0
3 years ago
Read 2 more answers
Mrs.Garcia used 5 cups of flour to make 3 loaves of bread. Many can she make if she only has 33 cups of flour?
Alex17521 [72]

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6 0
2 years ago
-1.84 - 2.3m= -2.3(m+.8)
alexira [117]

Answer:

m ∈ R

Step-by-step explanation:

- 1,84 - 2,3m = - 2,3(m + 0,8)

- 1,84 - 2,3m = - 2,3m - 1,84

m ∈ R

5 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
ipn [44]

Answer:

The mean of the distribution is about 53.9 and standard deviation is about 16.5 marks.

Step-by-step explanation:

Use the trick of transforming the given random variable one that is standard normal distributed, aka a z-score, then look up the two percentile values in z-score tables to get two equations with two unknowns.

So, let x be the variable describing the marks of a student, and

z=(x-\mu)/\sigma

the standardized equivalent of that (mu - mean, sigma - standard deviation).

We are looking for values of mu and sigma. At this point we'd be out of luck, but, wait, we're given two bits of info: the 10% point (aka, 90th percentile) and the 20% point (can be interpreted as 100-20th percentile). For each point we can use z-score tables to look up the corresponding values of z (just search for z tables). I found:

z-score for the 10% point: z_10=1.28

z-score for the 20% point: z_20=-0.84

That gives us two equations:

z_{10}=1.28=(75-\mu)/\sigma\\z_{20}=-0.84=(40-\mu)\sigma

and can be solved for mu and sigma (do the work on your end, I am showing my result):

\mu=53.87\,\,,\,\, \sigma=16.51

The mean of the distribution is about 53.9 and standard deviation is about 16.5 marks.

3 0
3 years ago
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