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3 years ago

5/3 is less than or greater than 5/2

2 answers:

7 0

These are improper fractions so you need to change them to mixed numbers. 5/2 is 2 1/2 and 5/3 is 1 2/3. From there you can tell which is smaller and which is larger.

oksano4ka [1.4K]3 years ago

5 0

5/2 is great .if u divid 5/3 it's 1.666 and 5/2 is 2.5 so 2.5 is great

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What 12+9 A 99 B 4 C 0

chubhunter [2.5K]

Answer:

d 21xysxhducux6cyvivivi

5 0

3 years ago

I need it done asapp Will give brainly!!

brilliants [131]

Answer:

Step-by-step explanation:

4(5-2)+(-7)(0.5)

20-8+(-3.5)

12-3.5

8.5

6 0

3 years ago

Read 2 more answers

Question Help

mina [271]

The answer is 44 and if he deposits it is 33 dollars:

11x4

1x4=4

So 11x4=44

And if he deposits 11 he will have 33

3 0

3 years ago

WILL GIVE BRAINLIEST NEED HELP

monitta

Hey there!

When two lines intersect, the pairs of angles that are directly across from each other (ex. the pair depicted in the image) are called opposite angles. These angles always measure the same.

This means that we can set the two expressions equal to each other and solve for x:

2x + 2 = 3x - 52

x = 54

Hope this helps!

8 0

3 years ago

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to interpret the confidence interval of (-20.5707, -11.4293), we can say...

We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.

5 0

2 years ago