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

What is 45 divided by 8.0

Alexxx [7]

Answer:

answer is 5.625

hope it helps

7 0

3 years ago

Read 2 more answers

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