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3 years ago

5

Nia plans to put enough money in the bank each week for 100 weeks so that the amount in the bank is always proportional to the w

eek number. On week 8, she had $20.00. On week 12, she had $30.00. How much money will be in the savings account on week 100?

2 answers:

dolphi86 [110]3 years ago

7 0

Answer:

40$

Step-by-step explanation:

because 30.00 and 20.00 equal 50$ and you would subtract the quations and get 40$

garri49 [273]3 years ago

3 0

Answer:

$40

Step-by-step explanation:

2 1/2 times the week

8 x 2.5 = 20

12 x 2.5 = 30

40 x 2.5 = 100

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For a particular diamond mine, 81% of the diamonds fail to qualify as "gemstone grade". A random sample of 92 diamonds is analyz

Fed [463]

Answer:

68.79% probability that more than 79% of the sample diamonds fail to qualify as gemstone grade

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 92, p = 0.81$

So

$\mu = E(X) = np = 92*0.81 = 74.52$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{92*0.81*0.19} = 3.76$

Find the probability that more than 79% of the sample diamonds fail to qualify as gemstone grade.

This is 1 subtracted by the pvalue of Z when X = 0.79*92 = 72.68. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{72.68 - 74.52}{3.76}$

$Z = -0.49$

$Z = -0.49$ has a pvalue of 0.3121

1 - 0.3121 = 0.6879

68.79% probability that more than 79% of the sample diamonds fail to qualify as gemstone grade

4 0

3 years ago

What is the volume of the composite figure?

Stells [14]

The right answer is

D.)

Hope this helps :)

4 0

3 years ago

Read 2 more answers

A straight line, L, is perpendicular to the straight line y=2x-1 and passes through the point (8,1). Find an equation of L

luda_lava [24]

Let's call that other line with the eqn. y=2x-1 line M.
This line is in slope-intercept form, which means that it is written in the form y=mx+b where m is the slope and b is the y-intercept.
This means that the slope of line M is 2.

Perpendicular lines have slopes which are opposite reciprocals.
(that is to say, if you flipped the fraction and changed the sign)

Of course, 2 isn't a fraction, but it's implied as 2/1.
The opposite reciprocal would then be -1/2.
Let's plug this into our slope-intercept form equation for line L.

y = mx + b
y = -1/2x + b

Of course, we still need to find that y-intercept. (y when x = 0)
To do this, we need to interpret the slope.
Slopes are rise over run, so m = -1/2 means a change of -1 in y = 2 in x.
Let's take a point we know is in our line, (8, 1).
To find that y-intercept, we want x to be 0.
To do this, we'd have to subtract 8 from x.
And according to our slope, this means adding 4 to y.
Our y-intercept is at (0, 5), with the value b that we use being just 5.

$\boxed{y=-\frac{1}2x+5}$

7 0

3 years ago

Use the following image to assist in your answer.<br> a =<br> - 6<br> - 9<br> - 4

vfiekz [6]

Step-by-step explanation:

Pythagoras' theorem for the smallest one :

$c^2 = 4^2 + 6^2$

$c^2 = 16 + 36$

$c^{2}$ = 52

Pythagoras' theorem for the middle one :

$b^{2}$ = $6^{2}$ + $a^{2}$

Pythagoras' theorem for the biggest one :

$(4+a)^2 = c^2 + b^2$

$16 + 8a + a^2 = 52 + b^2$

Using the formula before (for $b^2$ ) it becomes :

$16 + 8a + a^2 = 52 + (6^2 + a^2)$

$16 + 8a + a^2 = 52 + 6^2 + a^2$

$16 + 8a = 52 + 6^2$

16 + 8a = 52 + 36

16+8a = 88

8a = 88-16

8a = 72

$a = \frac{72}{8}$

a = 9

Verifying :

$b^2 = 6^2 + a^2$

$b^2 = 36 + 81$

$b^2 = 117$

$b^{2}$ = 117

The biggest one :

$(4+a)^2 = c^2 + b^2$

$(4+9)^2 = 52 + 117$

$13^2 = 169$

True

8 0

2 years ago

If c is 5, then 6x =

Firlakuza [10]

Answer:

30

Step-by-step explanation:

because if x is 5 then 6(5)

and 6x5=30

5 0

3 years ago