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3 years ago

Is 8:2 greater than 5:1? I need someone to help me clarify if this is correct. Thanks.

Mathematics

2 answers:

kaheart [24]3 years ago

3 0

Answer:

No, 8:2 = 4:1 which is less than 5:1

Step-by-step explanation:

svet-max [94.6K]3 years ago

3 0

Answer:

$\huge \fbox \pink {A}\huge \fbox \green {n}\huge \fbox \blue {s}\huge \fbox \red {w}\huge \fbox \purple {e}\huge \fbox \orange {r}$

$8 : 2 \\ = \frac{8}{2} \\ = 4$

$5 : 1 \\ = \frac{5}{1} \\ = 5$

8 : 2 = 4, while 5 : 1 = 5.

4 < 5.

•°• 8 : 2 is not greater than 5 : 1.

ʰᵒᵖᵉ ⁱᵗ ʰᵉˡᵖˢ

$\huge\blue{ \mid{ \underline{ \overline{ \tt ꧁❣ ʀᴀɪɴʙᴏᴡˢᵃˡᵗ2²2² ࿐ }} \mid}}$

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3 years ago

ABCD is a parallelogram. Find the measure of angle ABC.

iVinArrow [24]

Answer:

∠ABC = 50°

Step-by-step explanation:

Since this is a parallelogram, opposite angles are congruent.

Therefore, ∠D ≅ ∠B

∠D = 38° + 12°

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3 years ago

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When rolling two 6-sided cubes, what are the chances that the sum of the roll will be 7?

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3 years ago

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below

mafiozo [28]

Answer:

a) $\bar X = 369.62$

b) $Median=175$

c) $Mode =450$

With a frequency of 4

d) $MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

e) $s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

Replacing we got:

$\bar X = 369.62$

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

$Median=175$

Part c

The mode is the most repeated value in the sample and for this case is:

$Mode =450$

With a frequency of 4

Part d

The midrange for this case is defined as:

$MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

Part e

For this case we can calculate the deviation given by:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

5 0

3 years ago

HELP ME NOW>>>>>>>>>>>>>>>>

Nutka1998 [239]

Answer:

3.8

Step-by-step explanation:

1.8 + 3.6 = 5.4

5.4 - 9.2 = 3.8

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3 years ago