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soldi70 [24.7K]
3 years ago
13

A recipe calls for 3/4 cup of sugar per batch of cookies. If Gabe wants to make 6 batches of cookies, how many cups of sugar doe

s he need?
Mathematics
1 answer:
Gennadij [26K]3 years ago
3 0
\frac{3}{4} cups of sugar is used to make one batch of cookies.
So 6 batches would be 4 \frac{1}{2}.
\frac{3}{4}<span>×</span>6 = 4 \frac{1}{2} of sugar.
The answer is 4 1/2.
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Can u help me on this plzz
avanturin [10]
Median: 29
Range: 25
IQR: 14.5

Explanations:

**Median:**
(To find the median, we need to first order all the elements)

Ordered —> 15, 18, 18, 20, 23, 28, 30, 33, 33, 34, 38, 40

(Since there are an even number of elements, we need to add the two elements in the middle and divide by 2)

Median = (28 + 30)/2 = 58/2 = 29

**Range:**

(To find the range, you just have the subtract the smallest one from the largest)

Range = 40 - 15 = 25

**IQR:**

First half of elements —> 15, 18, 18, 20, 23, 28
Second half of elements —> 30, 33, 33, 34, 38, 40

Q1 (Quartile 1) = Median of first half = (18 + 20)/2 = 38/2 = 19
Q3 (Quartile 3) = Median of second half = (33 + 34)/2 = 67/2 = 33.5

IQR = Q3 - Q1 = 33.5 - 19 = 14.5
5 0
3 years ago
NO LINKS!!
tester [92]

Answer:

Step-by-step explanation:

6 0
3 years ago
Whoever gets this right gets brainlyess answer and it's worth 10 points :)
Lunna [17]
60% = 0.6
12 / 0.6 = 20

answer 12 is 60% of 20

8 0
3 years ago
A sign standing 6 feet tall is situated 18 feet away from a flag pole. at a certain time of day the sign's shadow is 3 feet long
alexdok [17]
Observe attached picture.

On picture we have:
A = height of flagpole = x ft
B = length of flagpole's shadow = 24 ft
C = height of sign = 6 ft
D = length of sign's shadow = 3 ft

When we draw a picture representing this problem we can also add another line marked in red. This way we can see that we have two right-angle triangles. We can see that both have same angle marked with α.

We can apply trigonometry rules to find height of flagpole.

From small triangle containing sign we can find tangens function:
tan \alpha = \frac{C}{D}
Similarly we can do for large triangle containing flagpole:
tan \alpha = \frac{A}{B}

We see that these two equations have same left sides. This means that their right sides must also be same:
\frac{C}{D} = \frac{A}{B}
We can solve for A:
CB=AD \\ A= \frac{CB}{D}  \\ A= \frac{6*24}{3}  \\ A=48 ft

Height of flagpole is 48 feet.

8 0
4 years ago
If the area around the vegetable garden is of uniform width (labeled with x) and the dimensions of the vegetable garden is 45 fe
Nady [450]
Area of Rectangle = Width × Length

Vegetable Garden
Width = 20
Length = 45
Area of Vegetable Garden = 20×45=900 Sq ft

Area of Entire Garden = Width × Length
Width = 20 + 2x
Length = 45 + 2x
Area = (20 + 2x)(45 + 2x)
(2x + 20)(2x + 45)
4x^2 + 90x + 45x + 900
Area of Entire Garden = 4x^2 + 135x + 900

Area of Flower Garden = Area of Entire Garden - Area of Vegetable Garden

(4x^2 + 135x + 900) - (20 × 45)
4x^2 + 135x + 900 - 900
Area of Flower Garden = 4x^2 + 135x

7 0
3 years ago
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