Which measurement is most accurate to describe the amount of water a teaspoon could hold?

Erwin had 2 5/12 gallons of lemonade. He spilled 11/12 gallon of lemonade. How much lemonade dose he have left? Write in simples

xz_007 [3.2K]

Answer:

Erwin have 1 $\frac{1}{2}$ gallons of lemonade left

Step-by-step explanation:

Erwin had 2 5/12 gallons of lemonade. He spilled 11/12 gallon of lemonade. How much lemonade dose he have left? Write in simplest form

All we need to do is to subtract 11/12 from 2 5/12. But to do this, we will convert 2 5/12 to improper fraction

converting 2 5/12 to improper fraction yields : 29/12

Lemonade have left = Total lemonade - lemonade spilled

= 29/12 - 11/12

= 18/12

Lemonade have left =18/12

But we can reduce 18/12 to its lowest term

18÷6/ 12÷6 = 3/2

Lemonade have left = 3/2

We can change to 3/2 to a mixed number to give us 1 $\frac{1}{2}$

Therefore Erwin have 1 $\frac{1}{2}$ gallons of lemonade left

6 0

4 years ago

Read 2 more answers

3. Solve. -11.0 – 3.5

Anon25 [30]

Answer:

-14.5

Step-by-step explanation:

Take - common: -(11.0+3.5)
Just add them in the bracket: -(14.5)
Remove the bracket: -14.5

5 0

3 years ago

Read 2 more answers

A researcher compares the effectiveness of two different instructional methods for teaching physiology. A sample of 178 students

xxTIMURxx [149]

Answer:

critical value = 1.645

The 90% confidence interval = ( -22.62, -17.58)

Step-by-step explanation:

Given that:

the sample size $n_1$ = 178

the sample size $n_2$ = 226

the sample mean $\overline x_1$ = 54.4

the sample mean $\overline x_2$ = 74.5

population standard deviation $\sigma_1$ = 18.58

population standard deviation $\sigma_2$ = 9.52

level of significance ∝ = 1 - 0.90 = 0.10

The critical value for $Z_{\alpha/2} = Z _{0.10/2} = Z_{0.005}$ is 1.645

For the construction of our confidence interval, we use 90% since that is used to find the critical value.

∴

The margin of error = $Z \times\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}$

$1.645 \times\sqrt{\dfrac{18.58^2}{178} + \dfrac{9.52^2}{226}}$

$1.645 \times\sqrt{\dfrac{345.2164}{178} + \dfrac{90.6304}{226}}$

$1.645 \times\sqrt{2.34042}$

$\simeq$ 2.52

The lower limit = $( \overline x_1 - \overline x_2) - (M.O.E)$

= ( 54.4-74.5) - (2.52)

= -20.1 - 2.52

= -22.62

The upper limit = $( \overline x_1 - \overline x_2) + (M.O.E)$

= ( 54.4-74.5) + (2.52)

= -20.1 + 2.52

= -17.58

The 90% confidence interval = ( -22.62, -17.58)

6 0

3 years ago

Help Needed ASAP (30 points)

Degger [83]

Answer:

D)

and

n=1

AD=7.07

Step-by-step explanation:

hope it helps

6 0

3 years ago

Item 10

Mazyrski [523]

Answer:

c=39

Step-by-step explanation:

c-7=32

c=7+32

c=39

4 0

3 years ago