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3 years ago

12

A patient drinks 5 12ounce cups of coffee daily. The doctor recommends cutting back 60%, how many ounces of coffee can the patie

nt have each day?

1 answer:

Dennis_Churaev [7]3 years ago

5 0

The patient can have 204.8 ounce cups of coffee every day.

Step-by-step explanation:

Initial consumption limit of the coffee by the patient- 512-ounce cups

Doctor recommendation-60% cut back in the coffee consumption

Thus according to the doctor's recommendation, if the patient needs to cut back 60% of the coffee consumption, then he is only to consume remaining (100-60)= 40% of his daily consumption.

Total daily consumption he can take= 40% of initial consumption

= (40/100) *512 ounces

= 204.8 ounces of coffee

Hence, the patient needs to take only 204.8 ounces of coffee complying by the Doctor’s recommendation.

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Because you said "estimate," I will be estimating this.

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3 years ago

30÷10 divide until you find the ten thousand digit

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yesterday at football practice, Han made 12 of his field goal attempts. Today he made 75% as many as he did yesterday. How many

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3 years ago

Read 2 more answers

If the average yield of cucumber acre is 800 kg, with a variance 1600 kg, and that the amount of the cucumber follows the normal

lorasvet [3.4K]

Answer:

a

The percentage is

$P(x_1 < X < x_2 ) = 51.1 \%$

b

The probability is $P(Z > 2.5 ) = 0.0062097$

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 800$

The variance is $var(x) = 1600 \ kg$

The range consider is $x_1 = 778 \ kg \ x_2 = 834 \ kg$

The value consider in second question is $x = 900 \ kg$

Generally the standard deviation is mathematically represented as

$\sigma = \sqrt{var (x)}$

substituting value

$\sigma = \sqrt{1600}$

$\sigma = 40$

The percentage of a cucumber give the crop amount between 778 and 834 kg is mathematically represented as

$P(x_1 < X < x_2 ) = P( \frac{x_1 - \mu }{\sigma} < \frac{X - \mu }{ \sigma } < \frac{x_2 - \mu }{\sigma } )$

Generally $\frac{X - \mu }{ \sigma } = Z (standardized \ value \ of \ X)$

So

$P(x_1 < X < x_2 ) = P( \frac{778 - 800 }{40} < Z< \frac{834 - 800 }{40 } )$

$P(x_1 < X < x_2 ) = P(z_2 < 0.85) - P(z_1 < -0.55)$

From the z-table the value for $P(z_1 < 0.85) = 0.80234$

and $P(z_1 < -0.55) = 0.29116$

So

$P(x_1 < X < x_2 ) = 0.80234 - 0.29116$

$P(x_1 < X < x_2 ) = 0.51118$

The percentage is

$P(x_1 < X < x_2 ) = 51.1 \%$

The probability of cucumber give the crop exceed 900 kg is mathematically represented as

$P(X > x ) = P(\frac{X - \mu }{\sigma } > \frac{x - \mu }{\sigma } )$

substituting values

$P(X > x ) = P( \frac{X - \mu }{\sigma } >\frac{900 - 800 }{40 } )$

$P(X > x ) = P(Z >2.5 )$

From the z-table the value for $P(Z > 2.5 ) = 0.0062097$

7 0

3 years ago

2. 3 qt =?L<br>15lb=?kg<br>7L=?qt

Firdavs [7]

3qt=2.83906L and 15lb=6.80389kg and 7L=7.39682qt hope this help

5 0

3 years ago

Read 2 more answers