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STALIN [3.7K]
3 years ago
15

How much of each ingredient will cassandra need to use in order to cut the recipe in half? the numbers to cut in half are 2 1\4

Cups, 2pkg., 3T., 1T., 1T., 6 Cups. Plz HELP
Mathematics
1 answer:
Oduvanchick [21]3 years ago
3 0
She will need to use 1 1/8 cup, 1pkg, 1 1/2 T, 1/2 T, 1/2 T and 3 cups. hope this helps.
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A car travels at an average speed of 50 km/h for 1.5 hours. How far did the car travel in km?
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Answer:

75 km

Step-by-step explanation:

(50km/h)(1.5h) =<em> 75km</em>

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True or false to find the percent of a number we convert to a fraction or a decimal and then divide the number
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Chef bought 3 3/4 kilograms of apples 7 1/4 kilograms of pears and 10 1/8 kilograms of oranges. How many kilograms of fruit is a
Hatshy [7]

Answer: 21 1/8 kilogram

Step-by-step explanation:

From the question, we are informed that Chef bought 3 3/4 kilograms of apples 7 1/4 kilograms of pears and 10 1/8 kilograms of oranges. The total kilogram of fruits bought would be:

= 3 3/4 + 7 1/4 + 10 1/8

Note that the Lowest Common Multiple is 8.

= 3 6/8 + 7 2/8 + 10 1/8

= 20 9/8

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6 0
3 years ago
The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly sel
garri49 [273]

Complete question:

The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly selects 45 customers and records the amount of time from the moment they stand in the back of a line until the moment the cashier scans their first item. He calculates the mean and standard deviation of this sample to be barx = 4.2 minutes and s = 2.0 minutes. If appropriate, find a 90% confidence interval for the true mean time (in minutes) that customers at this supermarket wait in a check-out line

Answer:

(3.699, 4.701)

Step-by-step explanation:

Given:

Sample size, n = 45

Sample mean, x' = 4.2

Standard deviation \sigma = 2.0

Required:

Find a 90% CI for true mean time

First find standard error using the formula:

S.E = \frac{\sigma}{\sqrt{n}}

= \frac{2}{\sqrt{45}}

= \frac{2}{6.7082}

SE = 0.298

Standard error = 0.298

Degrees of freedom, df = n - 1 = 45 - 1 = 44

To find t at 90% CI,df = 44:

Level of Significance α= 100% - 90% = 10% = 0.10

t_\alpha_/_2_, _d_f = t_0_._0_5_, _d_f_=_4_4 = 1.6802

Find margin of error using the formula:

M.E = S.E * t

M.E = 0.298 * 1.6802

M.E = 0.500938 ≈ 0.5009

Margin of error = 0.5009

Thus, 90% CI = sample mean ± Margin of error

Lower limit = 4.2 - 0.5009 = 3.699

Upper limit = 4.2 + 0.5009 = 4.7009 ≈ 4.701

Confidence Interval = (3.699, 4.701)

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8 The circle graph below shows the number of votes each candidate received in a recent student
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Answer:

60%

Step-by-step explanation:

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