All categories

3 years ago

I need help please!!!

Mathematics

1 answer:

nexus9112 [7]3 years ago

6 0

D) The measures of the non- right angles in each triangle .

You might be interested in

Literacy rates ~ Tamil Nadu is one of the 29 states in India, which is also one of the leaders in literacy initiatives carried o

Ivan

Answer:

Test statistic = 2.8965

Step-by-step explanation:

The test statistic is given by the formula

z = (x - μ)/σₓ

μ = mean = 0.799

x = proportion of literate people in Tamil Nadu = (983/1185) = 0.8295

σₓ = standard error of this calculated proportion = √[p(1-p)/n]

= √(0.8295 × 0.1705/1185) = 0.01093

z = (x - μ)/σₓ

z = (0.8295 - 0.799)/0.01093 = 2.8965

z = 2.8965

Hope this Helps!!!

4 0

3 years ago

Geometric Sequence S = 1.0011892 + ... + 1.0012 + 1.001 + 1

leva [86]

Answer:

$S_{1893} =5632.98$

Step-by-step explanation:

The correct form of the question is:

$S = 1.001^{1892} + ... + 1.001^2 + 1.001 + 1$

Required

Solve for Sum of the sequence

The above sequence represents sum of Geometric Sequence and will be solved using:

$S_n = \frac{a(1 - r^n)}{1 - r}$

But first, we need to get the number of terms in the sequence using:

$T_n = ar^{n-1}$

Where

$a = First\ Term$

$a = 1.001^{1892}$

$r = common\ ratio$

$r = \frac{1}{1.001}$

$T_n = Last\ Term$

$T_n = 1$

So, we have:

$T_n = ar^{n-1}$

$1 = 1.001^{1892} * (\frac{1}{1.001})^{n-1}$

Apply law of indices:

$1 = 1.001^{1892} * (1.001^{-1})^{n-1}$

$1 = 1.001^{1892} * (1.001)^{-n+1}$

Apply law of indices:

$1 = 1.001^{1892-n+1}$

$1 = 1.001^{1892+1-n}$

$1 = 1.001^{1893-n}$

Represent 1 as $1.001^0$

$1.001^0 = 1.001^{1893-n}$

They have the same base:

So, we have

$0 = 1893-n$

Solve for n

$n = 1893$

So, there are 1893 terms in the sequence given.

Solving further:

$S_n = \frac{a(1 - r^n)}{1 - r}$

Where

$a = 1.001^{1892}$

$r = \frac{1}{1.001}$

$n = 1893$

So, we have:

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{1 -\frac{1}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{1.001 -1}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001}^{1893})}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} *(1 -\frac{1}{1.001^{1893}})}{\frac{0.001}{1.001} }$

Simplify the numerator

$S_{1893} =\frac{1.001^{1892} -\frac{1.001^{1892}}{1.001^{1893}}}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} -1.001^{1892-1893}}{\frac{0.001}{1.001} }$

$S_{1893} =\frac{1.001^{1892} -1.001^{-1}}{\frac{0.001}{1.001} }$

$S_{1893} =(1.001^{1892} -1.001^{-1})/({\frac{0.001}{1.001} })$

$S_{1893} =(1.001^{1892} -1.001^{-1})*{\frac{1.001}{0.001}}$

$S_{1893} =\frac{(1.001^{1892} -1.001^{-1}) * 1.001}{0.001}$

Open Bracket

$S_{1893} =\frac{1.001^{1892}* 1.001 -1.001^{-1}* 1.001 }{0.001}$

$S_{1893} =\frac{1.001^{1892+1} -1.001^{-1+1}}{0.001}$

$S_{1893} =\frac{1.001^{1893} -1.001^{0}}{0.001}$

$S_{1893} =\frac{1.001^{1893} -1}{0.001}$

$S_{1893} =5632.97970294$

Hence, the sum of the sequence is:

$S_{1893} =5632.98$ ----- approximated

4 0

3 years ago

Samples of size n = 9 are selected from a population with μ = 80 with σ = 18. what is the expected value of m, the mean of the d

adoni [48]

Answer:
80

Explanation:
The mean of the distribution of sample means always target the population mean. What this line tells is that whatever the population mean will be, the mean of the distribution of sample means will be same. Since the population mean is 80, the mean of distribution of sample means will also be 80.

However, the standard deviation of the sample distributions is different and is equal to the population standard deviation divided by square root of sample size. So in this case the standard deviation of distribution of sample means will be 6 .

4 0

3 years ago

Can I have some help please

horrorfan [7]

Answer:

I think it's trapezium .

6 0

3 years ago

What is the answer to 2x^2 - 3x - 15 = 5

Alja [10]

2x²-3x-15=5
2x²-3x-15-5=0
2x²-3x-20=0
½(2x+5)(2x-8)

-2x=-5
x=-5/2

2x-8=0
2x=8
x=4

---------------------------

6 0

4 years ago