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3 years ago

Given that a = -2 and b = 7 ,evaluate the following algebraic expression (1) a(b2 -a)

Mathematics

2 answers:

marishachu [46]3 years ago

8 0

Answer:

-2(14--2)

-2*19

38.

I think this is the correct answer. Thanks

goldfiish [28.3K]3 years ago

5 0

Answer:

Answer is -32.

Step-by-step explanation:

a= -2

b=7

Expression: a(b2-a)

-2(7*2--2) (Place 'a' and 'b' values)

= -2(14+2) (Solve inside the bracket)

= -28-4 (Multiply -2 with the values inside the bracket)

= -32 (ans)

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Use the following information for the question. Male players at the high school, college and professional ranks use a regulation

Virty [35]

Answer:

13.567%

Step-by-step explanation:

We solve the above question, using z score formula

z = (x-μ)/σ, where

x is the raw score = 23.1 ounces

μ is the population mean = 22.0 ounces

σ is the population standard deviation = 1.0 ounce

More than = Greater than with the sign = >

Hence, for x > 23.1 ounces

z = 23.1 - 22.0/1.0

= 1.1

Probability value from Z-Table:

P(x<23.1) = 0.86433

P(x>23.1) = 1 - P(x<23.1)

P(x>23.1) = 1 - 0.86433

P(x>23.1) = 0.13567

Converting to percentage

= 0.13567 × 100

= 13.567%

Therefore, the percentage of regulation basketballs that weigh more than 23.1 ounces is 13.567%

8 0

2 years ago

Joe can cut and split a cord of firewood in 3 fewer hours than Dwight can. When they work together, it takes them 2 hours. How

aalyn [17]

Answer:

Dwight will take 6 hours to finish the job alone and Joe will take 3 hours to finish the job alone.

Step-by-step explanation:

Let us assume the time taken by Dwight to split a cord of firewood = K hrs

So, the per hour rate of Dwight = $(\frac{1}{K})$

As, Joe uses 3 LESS hours then Dwight.

So, the time taken by Joe to split a cord of firewood = (K- 3) hrs

So, the per hour rate of Joe = $(\frac{1}{K-3})$

Now, when both of them wok together, it takes them 2 hours.

So, the per hour rate of BOTH of them = $(\frac{1}{2})$

⇒ Per hour rate of ( Dwight + Joe) = $(\frac{1}{2} )$

$\implies (\frac{1}{K}) + (\frac{1}{K-3}) = (\frac{1}{2})$

Now, solving for the value of K , we get:

$(\frac{1}{K}) + (\frac{1}{K-3}) = (\frac{1}{2})\\\implies \frac{(K-3) + K}{K (K-3)} = (\frac{1}{2})\\\implies 2(2K -3) = K^2 - 3K\\\implies k^2 - 3K -4K +6 = 0\\\implies K^2 - 7K + 6= 0\\\implies K^2 - 6K - K + 6= 0\\\implies K(K-6) -1(K - 6)= 0\\\implies (K-6)(K-1) = 0$