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3 years ago

SIS HELP!! How many pennies are on the nth square?

Mathematics

2 answers:

anyanavicka [17]3 years ago

4 0

Answer:

B. , just did the assignment

ivann1987 [24]3 years ago

4 0

Answer:

The answer is B, 2^n - 1

Step-by-step explanation:

took the assignment on edge 2020

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Please help, seriously confused

Rufina [12.5K]

Original height = original width = x mm

1. She made an enlarged copy...
height = 2x
width = 2x

2. She cut off a rectangle...
height = 2x
width = $\frac{2}{3} *2x= \frac{4}{3}x$

3.She doubled the width...
height = 2x
width = $2*\frac{4}{3}x = \frac{8}{3}x$

height * width = area

$2x * \frac{8}{3}x=139 \ 968 \\ \\ \frac{16}{3}x^2= 139 \ 968 \\\\ x^2=139 \ 968: \frac{16}{3}=139 \ 968 * \frac{3}{16}= 8 \ 748*3 = 26 \ 244 \\ \\ x= \sqrt{26 \ 244} = 162 \ mm$

Original height was 162 mm

6 0

3 years ago

What are the like terms in this exspression 9a-3b + 4ab

vova2212 [387]

Answer:

none of these terms are like terms

7 0

3 years ago

Read 2 more answers

Solve for x. 3x+1=7 What is the root? If there is no root, choose none.

nikdorinn [45]

I think the root of "3x + 1 = 7" is 2...

6 0

3 years ago

Read 2 more answers

B) {(15.0). (15,-2)} Function? Explain:

Natalija [7]

No

Both share the same x-values which means it isn’t a function

Hope this helps ;)

5 0

3 years ago

2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi

Oksana_A [137]

Answer:

(a) $E(X) = 950$

(b) $COV = 0.007255$

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

$E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950$

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$COV = \frac{\sigma}{E(X)}$

Where the standard deviation is given by

$\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892$

So the coefficient of variance is

$COV = \frac{6.892}{950}$

$COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

$P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\$

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

$P(X > 980) = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980) = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980) = 1 - P(Z < 4.28)\\\\$

The z-score corresponding to 4.28 is 0.99999

$P(X > 980) = 1 - 0.99999\\\\P(X > 980) = 0.00001\\\\$

So it means that it is very unlikely that there will be more than 980 non-defective units.

8 0

4 years ago