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xz_007 [3.2K]
3 years ago
14

There are 105 calories in a glass and a half of lemonade. how many calories are in 2 1/2 glasses

Mathematics
2 answers:
ddd [48]3 years ago
8 0

Answer:

175

Step-by-step explanation:

A glass and a half is 1+1/2 = 3/2 glasses.

2 1/2 glasses is 2/2 glasses more than that, or 5/2 glasses.

5/2 glasses is 5/3 times the size of 3/2 glasses, so there will be 5/3 the number of calories:

... (105 calories) × (5/3) = 175 calories

lbvjy [14]3 years ago
5 0

Answer:

The answer is 175 calories in 2 1/2 glasses

Step-by-step explanation:

x = 2.5 x 105/1.5

x = 262.5/1.5

x = 175

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Suppose you have two similar rectangular prisms. The volume of the smaller rectangular prism is 125 in^3 and the volume of the l
Fittoniya [83]

Answer:

\boxed{ \frac{ 5}{ 9}}\\

Step-by-step explanation:

Data:

V₁ = 125 in³

V₂ = 729 in³

Calculations:

A scale factor (C) is the ratio of corresponding side lengths (l) of the two prisms.

C=\frac{ l_{1}}{ l_{2}}\\

The volume ratio is the cube of the scale factor.

\frac{V_{1}}{ V_{2}} = C^{3}\\

\frac{125}{ 729} = C^{3}\\

C = \sqrt [3]{\frac{ 125}{729 }} = \frac{5 }{9 }\\

\boxed{ \textbf{The scale factor is } \frac{ 5}{ 9}.}\\

7 0
3 years ago
Write an inequality for the given statement the product of -3 and a number is no more than 15
Nata [24]

Answer:

-3n ≤ 15

Step-by-step explanation:

"No more than 15" means the number can be less than 15 or equal to 15, but it can't be more than 15.

"The product of -3 and a number"  "is no more than"  "15".

                          -3n                                        ≤                 15

-3n ≤ 15

4 0
3 years ago
Anyone know this please ;)
PIT_PIT [208]

Answer:

The planet to have the biggest mass is Jupiter

the radius of mercury is 2.44 × 10^5

the difference in mass between Jupiter and Saturn is 2.2 × 10⁴

3 0
3 years ago
2. A marketing firm is trying to estimate the proportion of potential car buyers that would consider
Maurinko [17]

Answer:

a. The number of people that should be in the pilot study are 600 people

b. The point estimate is 0.62\overline 6

c. At 95% confidence level the true population proportion of potential car buyers of hybrid vehicle is between the confidence interval (0.588, 0.6654)

d. Two ways to reduce the margin of error are;

1) Reduce the confidence interval

2) Use a larger sample size

Step-by-step explanation:

a. The given parameters for the estimation of sample size is given as follows;

The margin of error for the confidence interval, E = 4% = 0.04

The confidence level = 95%

The sample size formula for a proportion as obtained from an online source is given as follows;

n = \dfrac{Z^2 \times P \times (1 - P)}{E^2}

Where, P is the estimated proportions of the desired statistic, therefore, we have for a new study, P = 0.5;

Z = The level of confidence at 95% = 1.96

n + The sample size

Therefore, we have;

n = \dfrac{1.96^2 \times 0.5 \times (1 - 0.5)}{0.04^2} = 600.25

Therefore, the number of people that should be in the pilot study in order to meet this goal at 95% confidence level is n = 600 people

b. The point estimate for the population proportion is the sample proportion  given as follows;

\hat p = \dfrac{x}{n}

Where;

x = The number of the statistic in the sample

n = The sample size

From the question, we have;

The number of potential car buyers, n = 600

The number of respondent in the sample that indicated that they would consider purchasing a hybrid, x = 376

Therefore, the point estimate, for the proportion of potential car buyers that would consider buying a hybrid vehicle, \hat p = 376/600 = 0.62\overline 6

c. The confidence interval for a proportion is given as follows

CI=\hat{p}\pm z\times \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

Therefore, we get;

CI=0.62 \overline 6\pm 1.96\times \sqrt{\dfrac{\hat{0.62 \overline 6}\cdot (1-\hat{0.62 \overline 6})}{600}}

C.I. ≈ 0.6267 ± 0.0387

The 95% confidence interval for the true population proportion of potential buyers of hybrid vehicle, C.I. =  (0.588, 0.6654)

d. The margin of error is given by the following formula;

MOE_\gamma = z_\gamma  \times \sqrt{\dfrac{\sigma ^2}{n} }

Where;

MOE_\gamma = Margin of error at a given level of confidence

z_\gamma = z-score

σ = The standard deviation

n = The sample size

Therefore, the margin error can be reduced by the following two ways;

1) Reducing the confidence interval and therefore, the z-score

2) Increasing the sample size

6 0
3 years ago
Help with his long question
lana [24]
4x³=756
x³=189
x=3 ³√7 should be acceptable
☺☺☺☺
3 0
3 years ago
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