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Lapatulllka [165]
3 years ago
9

5) How many thousands are there in 2 158 621?​

Mathematics
2 answers:
never [62]3 years ago
6 0

Answer:

2,158,621 = 2 (i think)

Sergio [31]3 years ago
3 0
2




Blah blah blah blah
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A town has a population of 32,000 in the year 2002; 38,400 in the year 2003; 46,080 in the year 2004; and 55,296 in the year 200
NeTakaya
The point is to find the growth rate. The compound formula is:

P=A(1+ growth rate)ⁿ, where A is the initial Value & P the new value after n years:
P₂₀₀₃ =P₂₀₀₂ (1+ growth rate)¹ (the period "n" from 2002 to 2003 being 1 year)

38400 = 32000(1+growth rate)¹
38400 / 32000 - 1= growth rate & growth rate = 1/5 = 0.2
You will balso find the same growth rate for:
P₂₀₀₄ = P₂₀₀₃(1+ growth rate)¹
P₂₀₀₅ = P₂₀₀₄((1+ growth rate)¹
between 2015 & 2002 THERE ARE 14 YEARS:

P₂₀₁₅ = P₂₀₀₂(1+0.2)¹⁴ & P₂₀₁₅ = 32000(1+02)¹⁴ = 410,854
8 0
4 years ago
How to do that? P p p p
klemol [59]

Answer:

Step-by-step explanation:

They tell you the answer in the picture. Angle DAB is the lower left vertex, so its measure is 80°

Angle BCD is the upper right vertex.

8 0
2 years ago
g The bursar a conducts a test of hypothesis that the mean amount of student debt a WCU undergrad will leave college with is equ
aleksandrvk [35]

Answer: Type I error

Step-by-step explanation:

A type 1 error is also referred to as the false positive and it is when a true null hypothesis is incorrectly rejected by the researcher.

On the other hand, type II error which is also refered to as the false-negative is when a null hypothesis which is false is failed to be rejected by the researcher but rather accepted.

Based on the information given in the question, since the data leads to the rejection of the null hypothesis such was later found out to be true, then a type I error has occured.

4 0
3 years ago
Which of the following functions functions has a rate of change that stays the same?
serious [3.7K]

Answer:

y = 2x   and y = -7x +9

Step-by-step explanation:

The two "linear" functions are those that have a constant rate of change. Actually, the slope (inclination) they show when plotted is that "rate of change".

The function : y = 2x, has a constant positive rate given by its slope (2).

The function : y= -7x +9 has a constant negative rate given by its slope (-7)

The other two functions are quadratic, represented by parabolas, and therefore don't have a constant rate of change.

8 0
3 years ago
Read 2 more answers
Consider the probability that no less than 92 out of 159 students will pass their college placement exams. Assume the probabilit
Yanka [14]

Answer:

0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.

Step-by-step explanation:

I am going to 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:

p = 0.55, n = 159. So

\mu = E(X) = 159*0.55 = 87.45

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{159*0.55*0.45} = 6.27

Probability that no less than 92 out of 159 students will pass their college placement exams.

No less than 92 is more than 91, which is 1 subtracted by the pvalue of Z when X = 91. So

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

Z = \frac{91 - 87.45}{6.27}

Z = 0.57

Z = 0.57 has a pvalue of 0.7157

1 - 0.7157 = 0.2843

0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.

5 0
3 years ago
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