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nadya68 [22]
3 years ago
9

Need help so baddddd

Mathematics
2 answers:
stepan [7]3 years ago
8 0
It would be 90%. 18 is 90% of 20
vampirchik [111]3 years ago
5 0

Answer:

90%

Step-by-step explanation:

18/ 20 = 18 × 5 20 × 5 = 90 100 = 90%.

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A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 48 specimens and counts the number of
baherus [9]

Answer: The open interval would be (31.4,42.5).

Step-by-step explanation:

Since we have given that

mean = 36.9

Standard deviation = 16.5

n = 48

At 98% confidence interval, z = 2.33

So, Interval would be

\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=36.9\pm 2.33\dfrac{16.5}{\sqrt{48}}\\\\=36.9\pm 5.549\\\\=(36.9-5.5,36.9+5.6\\\\=(31.4,42.5)

Hence, the open interval would be (31.4,42.5).

5 0
3 years ago
PLSSSSSS HELP ME WITH THISS!!!
ahrayia [7]
The answer is C ! The total would be $59
7 0
2 years ago
Read 2 more answers
Find the solution to the following: *
Trava [24]

Answer:

First one: x = all real numbers

Second one: x = 0

Step-by-step explanation:

for the first one

given 8x+10=2(4x+5) we need to isolate the variable (x) using inverse operations

step 1 distribute the 2 to what is in the parenthesis ( 4x and 5 )

2 * 4x = 8x

5 * 2 = 10

now we have 8x + 10 = 8x + 10

step 2 subtract 8x from each side

8x - 8x = 0

8x - 8x = 0

now we have 10 = 10

subtract 10 from each side

10 - 10 = 0

10 - 10 = 0

we're left with 0 = 0 meaning that all real numbers are solutions

For the second one

given 3x-8=2(x-4) once again we need to isolate the variable using inverse operations

step 1 distribute the 2 to what is in the parenthesis (x - 4)

2 *x = 2x

2 * -4 = -8

now we have 3x - 8 = 2x - 8

step 2 add 8 to each side

-8 + 8 = 0

-8 + 8 = 0

now we have 2x = 3x

step 3 subtract 2x from each side

3x - 2x = x

2x - 2x = 0

we're left with x = 0

5 0
3 years ago
Find the value of f(5) y = f(x) on a graph
Eva8 [605]

graphing

f ( 5 )

y =  f ( x )

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3 years ago
A spinner with
CaHeK987 [17]
1/6 if there are 6 different colors
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3 years ago
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