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8_murik_8 [283]

3 years ago

15

Lindsey made 4 lunches out of 1\2 of a leftover casserole she put the same amount in each lunch.how much of a casserole did she

put on each lunch

2 answers:

Andrei [34K]3 years ago

7 0

Answer:

1/8 because 1/2 divided by 1/2 is 1/4 and 1/4 divided by 1/4 is 1/8 and that's four lunches

Step-by-step explanation:

irakobra [83]3 years ago

3 0

Answer:

The answer is 1/8.

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A standard brick is 3.625 inches wide, 2.25 inches high, and 7.625 inches long. How many bricks could be in a pallet of bricks t

Snezhnost [94]

Answer: Option D. 1,778

Solution:

Standard brick:

Width: w=3.625 in

Height: h=2.25 in

Length: l=7.625 in

Volumen of one standard brick: v

v=w*h*l

v=(3.625 in)*(2.25 in)*(7.625 in)

v=62.19140625 in^3

Pallet of bricks:

Side: s=4 feet

s=(4 feet)*(12 in / 1 feet)→s=48 in

Volume of a pallet of bricks: V=s^3

V=(48 in)^3

V=110,592 in^3

Number of bricks could be in a pallet: n

n=V/v

n=(110,592 in^3) / (62.19140625 in^3)

n=1,778.252119

n=1,778

6 0

2 years ago

Read 2 more answers

The function f is defined by f(x) = 2x²-1.<br> Find f (2x).

umka21 [38]

Answer: $f(2x)=8x^2-1$

Step-by-step explanation:

All that is required here is to substitute x = 2x into the right side in the same way that is done for a numeric value.

$f(2x) = 2(2x)^2-1\\f(2x) = 2(4)x^2-1\\f(2x) = 8x^2-1\\$

5 0

2 years ago

Aproximati prin lipsa si prin adaos la mii numerele:

siniylev [52]

Do you talk romanian. and i think that it is g.

6 0

2 years ago

Graph the functions and approximate an x-value in which the exponential function exceeds the polynomial function. y = 4x y = 7x2

nordsb [41]

Answer:

The Exponential function is

$y=4^x$

And the polynomial function is

$y=7 x^2 +4 x -2$

And, we have to find the value of x for which, exponential function exceeds the polynomial function which can be written as

$4^x> 7 x^2 +4 x -2$

1. When , x= -1

LHS

$4^{-1}=\frac{1}{4}=0.25$

RHS

$=7*(-1)^2 +4 *(-1)-2\\\\=7-4-2=1$

2. When , x=0

L HS

$4^0=1$

RHS

$7*0+4*0-2= -2$

3. When ,x= 0.5

L HS

$4^{0.5}=2$

RHS

$=7*0.25 +4*0.5 -2\\\\= 1.75+2-2\\\\=1.75$

4. When , x=2

LHS

$4^{2}=16$

RHS

$=7*4^2+4*4-2\\\\=112+16-2\\\\=126$

The Minimum value for which exponential function exceeds the polynomial function is , x= 0.5

But,there is other value for which exponential function exceeds the polynomial function is , x=2.

5 0

2 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

2 years ago