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ddd [48]
3 years ago
12

How many dimensions does a plane have? A. TWO B. Zero c. Three D.One

Mathematics
1 answer:
Art [367]3 years ago
7 0

Answer:

3

Step-by-step explanation:

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What is 45 divided by 8.0
Alexxx [7]

Answer:

answer is 5.625

hope it helps

7 0
3 years ago
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How many cupcake are in 1/3 dozen
Anettt [7]

4 cupcakes because if you do 12/ 1/.. you do kcf = to 12 x 1/3= 4

your welcome

7 0
3 years ago
Write the linear equation in slope intercept form 2x+By=16
allochka39001 [22]

Answer:

Part 1) y=-\frac{2x}{B}x+\frac{16}{B}

Part 2) y=-\frac{1}{4}x+2

Step-by-step explanation:

Part 1)

we know that

The equation of the line in slope intercept form is equal to

y=mx+b

we have

2x+By=16

Isolate the variable y

subtract 2x both sides

By=-2x+16

Divide by B both sides

y=-\frac{2x}{B}x+\frac{16}{B}

Part 2)

we know that

The equation of the line in slope intercept form is equal to

y=mx+b

we have

2x+8y=16

Isolate the variable y

subtract 2x both sides

8y=-2x+16

Divide by 8 both sides

y=-\frac{2x}{8}x+\frac{16}{8}

Simplify

y=-\frac{1}{4}x+2

3 0
3 years ago
Is (x+4)^2equivalent to 2x^2+8x+8
Mashcka [7]

Answer:

(x+4)^{2} = x^{2}+8x+16

Step-by-step explanation:

5 0
3 years ago
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73
faust18 [17]

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62      [from the standard t table]

Therefore,

CI = (Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})

Thus,

Lower limit of CI =  (Mean - \frac{t\times\sigma}{\sqrt{n}})

or

Lower limit of CI =  (98.3 - \frac{2.62\times0.73}{\sqrt{103}})

or

Lower limit of CI = 98.11

and,

Upper limit of CI =  (Mean + \frac{t\times\sigma}{\sqrt{n}})

or

Upper limit of CI =  (98.3 + \frac{2.62\times0.73}{\sqrt{103}})

or

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and  the mean temperature could very possibly be 98.6°F

7 0
3 years ago
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