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2 years ago

3b - c +5 if c= - 4, b=2.

Mathematics

1 answer:

AnnZ [28]2 years ago

3 0

Answer:

Step-by-step explanation:

To find the answer, plug the values of each variable into the equation and simplify. It may be easiest to plug one variable in at a time, simplify, then plug the next variable in. Remember, if the equation is asking you to subtract a negative number, the negatives cancel and you want to add the numbers.

b = 2 c = -4

3b - c + 5 <----- Original expression

3(2) - c + 5 <----- Plug 2 in for "b"

6 - c + 5 <----- Multiply 3 and 2

11 - c <----- Add 6 and 5

11 - (-4) <----- Plug -4 in for "c"

15 <----- Add 11 and 4

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Which expression is equivalent to 3/2•3/2•3/2•3/2?

serg [7]

The answer is A) 2/3 ^ 4 power

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3 years ago

Suppose an airline policy states that all baggage must be box shaped with a sum of length, width, and height not exceeding 114

NISA [10]

Answer:

Step-by-step explanation:

Represent the length of one side of the base be s and the height by h. Then the volume of the box is V = s^2*h; this is to be maximized.

The constraints are as follows: 2s + h = 114 in. Solving for h, we get 114 - 2s = h.

Substituting 114 - 2s for h in the volume formula, we obtain:

V = s^2*(114 - 2s), or V = 114s^2 - 2s^3, or V = 2*(s^2)(57 - s)

This is to be maximized. To accomplish this, find the first derivative of this formula for V, set the result equal to 0 and solve for s:

----- = 2[(s^2)(-1) + (57 - s)(2s)] = 0 = 2s^2(-1) + 114s - 2s^2

Simplifying this, we get dV/ds = -4s^2 + 114s = 0. Then either s = 28.5 or s = 0.

Then the area of the base is 28.5^2 in^2 and the height is 114 - 2(28.5) = 57 in

and the volume is V = s^2(h) = 46,298.25 in^3

7 0

3 years ago

What's the value of 6x to the second power +4x+8, when x=7

lidiya [134]

6(7)^2+4(7)+8=y
6(49)+28+8=y
294+28+8=y
330=y

7 0

3 years ago

Read 2 more answers

The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir

Paha777 [63]

Answer:

95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean time = 5.15 years

$\sigma$ = sample standard deviation = 1.68 years

n = sample of college graduates = 4400

$\mu$ = population mean time

Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $5.15-1.96 \times {\frac{1.68}{\sqrt{4400} } }$ , $5.15+1.96 \times {\frac{1.68}{\sqrt{4400} } }$ ]

= [5.10 , 5.20]

Therefore, 95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

8 0

3 years ago

50+51+52+...+101=... Can you help me? I do not understand

Dmitriy789 [7]

Suppose

$S=50+51+52+\cdots+100+101$

At the same time, we can write

$S^*=101+100+99+\cdots+51+50$

Note that $S=S^*$ (just reverse the sum). Let's pair the first terms of $S$ and $S^*$ , and the second, and the third, and so on:

$S+S^*=(50+101)+(51+100)+(52+99)+\cdots+(100+51)+(101+50)$

Now, each grouped term in the sum on the right side adds to 151. There are 52 grouped terms on that same side (because there are 50 numbers in the range of integers 51-100, plus 50 and 101), which menas

$S+S^*=52\cdot151$

But $S=S^*$ , as we pointed out, so

$2S=52\cdot151\implies S=\dfrac{52\cdot151}2=3926$

5 0

2 years ago