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max2010maxim [7]
3 years ago
13

Solve. You must factor first. 8n^2 29n + 15 = 0 Please show work

Mathematics
1 answer:
lord [1]3 years ago
8 0

Answer: Multiply n299 by n by adding the exponents

Step-by-step explanation:

8n230+15=0

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From 1929 through the early 1930s, the prices of consumer goods actually decreased. Economists call this phenomenon deflation. T
ludmilkaskok [199]

Answer:

23.42

Step-by-step explanation:

If something is decreasing by 7% we can mulitply it by (1-.07) or .93

so our formula is

100(.93)²⁰= 23.42388737

which rounds to 23.42

4 0
3 years ago
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A SOLUTION CONTAINS 60% WATER, 15% BLEACH AND THE
Allushta [10]

Answer:

20 ounces of cleaner

Step-by-step explanation:

Water 48 oz is 60% of unknown total. 60% = 3/5. Create equation and solve

48= 3x/5. 240= 3x. 80 = x

Total of water, bleach and cleaner is 80 ounces

60% + 15% = 75%

100% - 75%= 25% cleaner

.25 × 80 = 20 ounces

6 0
3 years ago
Write a fact with the same sum as 7+ 5.
lisov135 [29]
0+12, 11+1, 4x3, 6x2, etc.
6 0
3 years ago
Read 2 more answers
From 5 a.m. to noon, the temperature rose 15 °F to a high of 10 °F.
kodGreya [7K]

Answer:

The temperature was 5 °F

7 0
3 years ago
We have n = 100 many random variables Xi ’s, where the Xi ’s are independent and identically distributed Bernoulli random variab
777dan777 [17]

Answer:

(a) The distribution of X=\sum\limits^{n}_{i=1}{X_{i}} is a Binomial distribution.

(b) The sampling distribution of the sample mean will be approximately normal.

(c) The value of P(\bar X>0.50) is 0.50.

Step-by-step explanation:

It is provided that random variables X_{i} are independent and identically distributed Bernoulli random variables with <em>p</em> = 0.50.

The random sample selected is of size, <em>n</em> = 100.

(a)

Theorem:

Let X_{1},\ X_{2},\ X_{3},...\ X_{n} be independent Bernoulli random variables, each with parameter <em>p</em>, then the sum of of thee random variables, X=X_{1}+X_{2}+X_{3}...+X_{n} is a Binomial random variable with parameter <em>n</em> and <em>p</em>.

Thus, the distribution of X=\sum\limits^{n}_{i=1}{X_{i}} is a Binomial distribution.

(b)

According to the Central Limit Theorem if we have an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.  

The sample size is large, i.e. <em>n</em> = 100 > 30.

So, the sampling distribution of the sample mean will be approximately normal.

The mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu=p=0.50

And the standard deviation of the distribution of sample mean is given by,

\sigma_{\bar x}=\sqrt{\frac{\sigma^{2}}{n}}=\sqrt{\frac{p(1-p)}{n}}=0.05

(c)

Compute the value of P(\bar X>0.50) as follows:

P(\bar X>0.50)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{0.50-0.50}{0.05})\\

                    =P(Z>0)\\=1-P(Z

*Use a <em>z</em>-table.

Thus, the value of P(\bar X>0.50) is 0.50.

8 0
3 years ago
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