The prime factorization of 2020 is 2² × 5¹ × 101¹. How many different whole numbers are factors of 2020?

What is the least common denominator of 1/3 and 5/9

crimeas [40]

Answer:

Equivalent Fractions with the LCD

1/3 = 3/9

5/9 = 5/9

Solution:

Rewriting input as fractions if necessary:

1/3, 5/9

For the denominators (3, 9) the least common multiple (LCM) is 9.

LCM(3, 9)

Therefore, the least common denominator (LCD) is 9.

Calculations to rewrite the original inputs as equivalent fractions with the LCD:

1/3 = 1/3 × 3/3 = 3/9

5/9 = 5/9 × 1/1 = 5/9

4 0

3 years ago

You have 20 bucks in your pocket.

diamong [38]

Answer:

the answer would be $45.55

<3

Step-by-step explanation:

20$ + $50.00 = $70

$70 - $6.95 = $64.5

$64.5 - $9.95 = $54.55

$54.55 - $20 = $34.55

$34.55 + $15 = $49.55

$49.55 - $4 = $45.55

5 0

3 years ago

96 ÷ 42 + (25 x 2) - 15 - 3=? I need help with my math

ANTONII [103]

Pemdas
so do parenthaseese first
work inside then out
(25 times 2)
(50)
now we have
96/42+(50)-15-3=?

start from the begining again of the equation (left) now we do division
96/42=16/7

now we have 16/7+(50)-15-3
start over and do addition
16/7+50=2 and 2/7+50=2+2/7+50=52 and 2/7
subtract
52 and 2/7-15=52+2/7-15=37 +2/7=37 and 2/7
then subtract 3
37 and 2/7-3=37+2/7-3=34+2/7=34 and 2/7

answer is 34 and 2/7

7 0

3 years ago

Two statistics teachers both believe that each has the smarter class. To put this to the test, they give the same final exam to

SVETLANKA909090 [29]

Answer:

We conclude that there is no difference between the two classes.

Step-by-step explanation:

We are given that two statistics teachers both believe that each has a smarter class.

A summary of the class sizes, class means, and standard deviations is given below:n1 = 47, x-bar1 = 84.4, s1 = 18n2 = 50, x-bar2 = 82.9, s2 = 17

Let $\mu_1$ = mean age of student cars.

$\mu_2$ = mean age of faculty cars.

So, Null Hypothesis, $H_0$ : $\mu_1=\mu_2$ {means that there is no difference in the two classes}

Alternate Hypothesis, $H_A$ : $\mu_1\neq \mu_2$ {means that there is a difference in the two classes}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean age of student cars = 8 years

$\bar X_2$ = sample mean age of faculty cars = 5.3 years

$s_1$ = sample standard deviation of student cars = 3.6 years

$s_2$ = sample standard deviation of student cars = 3.7 years

$n_1$ = sample of student cars = 110

$n_2$ = sample of faculty cars = 75

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(47-1)\times 18^{2}+(50-1)\times 17^{2} }{47+50-2} }$ = 17.491

So, the test statistics = $\frac{(84.4-82.9)-(0)}{17.491 \times \sqrt{\frac{1}{47}+\frac{1}{50} } }$ ~ $t_9_5$

= 0.422

The value of t-test statistics is 0.422.

Now, the P-value of the test statistics is given by;

P-value = P( $t_9_5$ > 0.422) = From the t table it is clear that the P-value will lie somewhere between 40% and 30%.

Since the P-value of our test statistics is way more than the level of significance of 0.04, so we have insufficient evidence to reject our null hypothesis as our test statistics will not fall in the rejection region.

Therefore, we conclude that there is no difference between the two classes.

6 0

3 years ago

After mini-golfing, Alex and Mel meet up with their friends Promise and Zion for ice cream. They each get a single scoop of ice

Nitella [24]

Answer:

22 cups

Step-by-step explanation:

who names someone Promise

6 0

3 years ago