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Stels [109]
3 years ago
15

How many fluid ounces are in 3 pints

Mathematics
2 answers:
Vesnalui [34]3 years ago
8 0

There are 48 fluid ounces.

AysviL [449]3 years ago
5 0

48 fluid ounces = 3 pints

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Someone pls help meee plssss fast
Kryger [21]

Answer: choices 1, 4, and 5.

Step-by-step explanation: uhhhh the answers basically explain why. 1 because it's a terminating decimal, 4 because it's a repeating number, and 5 because the square root of 11 is not a perfect square, and is not a repeating or terminating decimal.

7 0
3 years ago
Help is much appreciated x
RoseWind [281]

Answer:

23.2 cm

Step-by-step explanation:

For the angle of 53.2 deg, x is the opposite leg. The hypotenuse is 29 cm.

sin A = opp/hyp

sin 53.2 deg = x/29 cm

x = 29 cm * sin 53.2 deg

x = 23.2 cm

6 0
3 years ago
Help pleaseeeeeeeeeeeeeee
TEA [102]

Answer:

54m2

Step-by-step explanation:

Cut the shape into two - from the 9m length. One has 3m by 8m(ie 5m + 3m) dimensions, and the other 6m(ie 9m - 3m) by 5m dimensions.

Taking area of first one=24m2,then area of the second =30m2.

To get area of whole shape, we add both areas=54m2

NB:Area=l×b

4 0
3 years ago
Read 2 more answers
The diameters of ball bearings are distributed normally. The mean diameter is 52 millimeters and the standard deviation is 4 mil
tatyana61 [14]

Answer:

0.1057

Step-by-step explanation:

We solve using z score formula.

z = (x-μ)/σ, where

x is the raw score = 57mm

μ is the population mean = 52mm

σ is the population standard deviation = 4mm

z = 57 - 52/4

z = 1.25

Probability value from Z-Table:

P(x<57) = 0.89435

P(x>57) = 1 - P(x<57)

1 - 0.89435

= 0.10565

Approximately = 0.1057

The probability that the diameter of a selected bearing is greater than 57 millimeters is 0.1057

6 0
3 years ago
Professor Melendez has 10 students in her college
frez [133]

Answer:

Mean = 18.625

Median = 18.5

Step-by-step explanation:

Given the following :

Age = 19, 27, 19, 18, 18, 18, 47, 19, 20, 18

The mean Age :

Mean = Σ(x) ÷ N

N = number of data

(19 + 27 + 19 + 18 + 18 + 18 + 47 + 19 +20 + 18) / 10

Σ(x) / N = 223/10 = 22.3

Rearranging to get the median:

18,18,18,18,19,19,19,20,27,47

Middle value = median ( 19 + 19)/2 = 19

The outliers in the data are:

18,18,18,18,19,19,19,20,27,47

OUTLIERS ARE VALUES

< Q1 - (1.5 × IQR)

> Q3 + (1.5 × IQR)

IQR = Q3 - Q1 (Interquartile range)

Q3 = upper quartile

Q1 = lower quartile

From the data :

Q3 = 20, Q1 = 18

IQR = Q3 - Q1 = 20 - 18 = 2

< 18 - (1.5 × 2) ; <15

> 20 + (1.5 × 2) ; >23

VAlues greater than 23 and less than 15 are outliers in the data

27 and 47

After removing outliers

N = 10 - 2 = 8

Σ(x) = 223 - (47+27) = 149

Mean = 149/8 = 18.625

X = 18,18,18,18,19,19,19,20

Median = (18 + 19)/2 = 18.5

3 0
4 years ago
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