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3 years ago

Simplify the expression. (–h4)5

Mathematics

2 answers:

sveta [45]3 years ago

4 0

-20h

unless i'm missing something.

katrin [286]3 years ago

4 0

Answer:

-h^20

Hope this help ! !

{ If so mark brainliest I'd appreciate it ;) }

- Nahannah Muhammad

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An automobile repair shop has determined that the average servic etime on an automobile is 2 hours with a standard deviation of

TiliK225 [7]

avg. service time = 2hrs +/- 32 mins = 180 mins +/- 32 mins = 148 mins or 212 mins.

Random sample = 64/400 = 16 % of the total population.

Seriously I dont know the answer to the problem, but it is an interesting question, and it forces me to study probability more.

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3 years ago

What is the equation of the bold line on the coordinate grid?

padilas [110]

.....the answer is y=3

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3 years ago

If C (x) = 65x + 850 and R(x) = 100x<br> find: C (500)

natali 33 [55]

Step-by-step explanation:

C(500)=65×500+850

=32500+850

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The required answer

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3 years ago

Convert the recursive equation that is below into an explicit equation where a is the initial value and the time elapsed is 1 ye

aleksley [76]

Answer:

f(n) = a+200(n -1)

Step-by-step explanation:

The constant difference between terms indicates the sequence is an arithmetic one. The explicit formula for an arithmetic sequence is ...

an = a1 + d(n -1)

where a1 is the first term and d is the common difference.

Your first term is "a", and your common difference is 200, so the n-th term of the sequence is ...

an = a + 200(n -1)

Written as a function of n, this is ...

f(n) = a + 200(n -1)

_____

Based on the problem description, we cannot tell how n relates to time, so we have created an f(n) that gives the same result as the recursive definition of f(n).

8 0

3 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Anna [14]

Answer:

a) 0.4452

b) 0.0548

c) 0.0501

d) 0.9145

e) 6.08 minutes or greater

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.7 minutes

Standard Deviation, σ = 0.50 minutes.

We are given that the distribution of length of the calls is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(calls last between 4.7 and 5.5 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{4.7 - 4.7}{0.50} \leq z \leq \displaystyle\frac{5.5-4.7}{0.50}) = P(0 \leq z \leq 1.6)\\\\= P(z \leq 1.6) - P(z$

$P(4.7 \leq x \leq 5.5) = 44.52\%$

b) P(calls last more than 5.5 minutes)

$P(x > 5.5) = P(z > \displaystyle\frac{5.5-4.7}{0.50}) = P(z > 1.6)\\\\P( z > 1.6) = 1 - P(z \leq 1.6)$

Calculating the value from the standard normal table we have,

$1 - 0.9452 = 0.0548 = 5.48\%\\P( x > 5.5) = 5.48\%$

c) P( calls last between 5.5 and 6 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{5.5 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(1.6 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$

$P(5.5 \leq x \leq 6) = 5.01\%$

d) P( calls last between 4 and 6 minutes)

$P(4 \leq x \leq 6) = P(\displaystyle\frac{4 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(-1.4 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$