If one inch represents 65 miles on a map, how many inches will represent 1,885 miles?

Use the approach in Gauss's problem to find the sums of the arithmetic sequences below

I am Lyosha [343]

Answer:

15150 is the required sum.

Step-by-step explanation:

Gauss method of summing sequences like this is to add the first term to the last term, the second term to the second to the last term and so on until you get to the middle. By doing this, you will always get the same sum. Then you divide the number of terms by two, because you are technically adding the first half of the sequence to the second half. You then multiply the sum of each pair that gave the same number with the number of terms that has been divided by 2. For this problem we have:

3 + 300 = 303

6 + 297 = 303

9 + 294 = 303

and so on.

100÷ 2 = 50.

Therefore we have 50 * 303 = 15150 is the required sum.

5 0

3 years ago

1.44 Make-up exam: In a class of 28 students, 27 of them took an exam in class and 1 student took a make-up exam the following d

Olenka [21]

Answer:

a) Decrease

b) New mean = 78.43

c) Decrease

Step-by-step explanation:

We are given the following in the question:

Total number of students in class = 28

Average of 27 students = 79

Standard Deviation of 27 students = 6.5

New student's score = 63

a) The new student's score will decrease the average.

b) New mean

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean = \dfrac{\displaystyle\sum x_i}{27} = 79\\\\\sum x_i = 27\times 79 = 2133$

New mean =

$\text{ New mean} =\dfrac{ \displaystyle\sum x_i +63}{28}\\\\ =\dfrac{2133+63}{28}= \dfrac{2196}{28} = 78.43$

Thus, the new mean is 78.43

c) Since the new mean decreases, standard deviation for new scores will decrease.

This is because the new value is within the usual values i.e. within two standard deviations of the mean. So, this wont cause a lot of variation as this value will be closer to already available data values. Also number of observations (n) in the denominator is increasing. Based on both these points we can conclude that standard deviation will decrease

Formula for Standard Deviation:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

7 0

3 years ago

What is an equation of the line that passes through the points (-3,-1) and (-4, –4)?

olga55 [171]

Answer:

y = 3x + 8

Step-by-step explanation:

6 0

3 years ago

Which expression is a difference of cubes? 9w^33-y^12 18p^15-q^21 36a^22-b^16 64c^15- a^26

LiRa [457]

we know that

A polynomial in the form $a^{3}-b^{3}$ is called adifference of cubes. Both terms must be a perfect cubes

Let's verify each case to determine the solution to the problem

case A) $9w^{33} -y^{12}$

we know that

$9=3^{2}$ ------> the term is not a perfect cube

$w^{33}=(w^{11})^{3}$ ------> the term is a perfect cube

$y^{12}=(y^{4})^{3}$ ------> the term is a perfect cube

therefore

The expression $9w^{33} -y^{12}$ is not a difference of cubes because the term $9$ is not a perfect cube

case B) $18p^{15} -q^{21}$

we know that

$18=2*3^{2}$ ------> the term is not a perfect cube

$p^{15}=(p^{5})^{3}$ ------> the term is a perfect cube

$q^{21}=(q^{7})^{3}$ ------> the term is a perfect cube

therefore

The expression $18p^{15} -q^{21}$ is not a difference of cubes because the term $18$ is not a perfect cube

case C) $36a^{22} -b^{16}$

we know that

$36=2^{2}*3^{2}$ ------> the term is not a perfect cube

$a^{22}$ ------> the term is not a perfect cube

$b^{16}$ ------> the term is not a perfect cube

therefore

The expression $36a^{22} -b^{16}$ is not a difference of cubes because all terms are not perfect cubes

case D) $64c^{15} -a^{26}$

we know that

$64=2^{6}=(2^{2})^{3}$ ------> the term is a perfect cube

$c^{15}=(c^{5})^{3}$ ------> the term is a perfect cube

$a^{26}$ ------> the term is not a perfect cube

therefore

The expression $64c^{15} -a^{26}$ is not a difference of cubes because the term $a^{26}$ is not a perfect cube

I'm adding a new case so I can better explain the problem

case E) $64c^{15} -d^{27}$

we know that

$64=2^{6}=(2^{2})^{3}$ ------> the term is a perfect cube

$c^{15}=(c^{5})^{3}$ ------> the term is a perfect cube

$d^{27}=(d^{9})^{3}$ ------> the term is a perfect cube

Substitute

$64c^{15} -d^{27}=((2^{2})(c^{5}))^{3}-(d^{9})^{3}$

therefore

The expression $64c^{15} -d^{27}$ is a difference of cubes because all terms are perfect cubes

5 0

3 years ago

Read 2 more answers

Can someone help with step by step

MrMuchimi

Answer:

See explanation

Step-by-step explanation:

$In\: \triangle ABC \:\&\: \triangle EDC\\\angle ABC \cong \angle EDC... (Right \: \angle s) \\\angle ACB \cong \angle ECD... (Vertical \: \angle s) \\\therefore \triangle ABC \sim \triangle EDC... (By\: AA\: Postulate) \\$

3 0

3 years ago