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aev [14]
3 years ago
9

A recipe calls for 5 deciliters of water and 236 milliliters of milk. How many milliliters of liquid does the recipe call for in

total?
Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
3 0
1 deciliter is one-tenth of a liter and 1 milliliter is one-thousandth of a liter.
So: 1 liter = 10 deciliter = 100 centiliter = 1000 milliliter

This means that 1 deciliter is equal to 100 milliliters.
Based on the above:

5 deciliters = 500 milliliters
500+236 = 736

The recipe calls for 736 milliliters of liquid in total.
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HELP Eva works different hours each week. These figures show her net pay for the last 6 weeks.
sleet_krkn [62]
To do this, we must find the mean of the first four weeks in the given weeks pay
So 218.52+388.67+317.68+392.11=1316.98
1316.98/4= 329.24
This is the answer and Hope this helps
5 0
3 years ago
Suppose a geyser has a mean time between eruptions of 72 minutes. Let the interval of time between the eruptions be normally dis
nikitadnepr [17]

Answer:

(a) The probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is 0.3336.

(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 ​minutes is 0.0582.

(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72​, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

(a) Let X = <u><em>the interval of time between the eruptions</em></u>

So, X ~ N(\mu=72, \sigma^{2} =23^{2})

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is given by = P(X > 82 min)

       P(X > 82 min) = P( \frac{X-\mu}{\sigma} > \frac{82-72}{23} ) = P(Z > 0.43) = 1 - P(Z \leq 0.43)

                                                           = 1 - 0.6664 = <u>0.3336</u>

The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{13} } } ) = P(Z > 1.57) = 1 - P(Z \leq 1.57)

                                                           = 1 - 0.9418 = <u>0.0582</u>

The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

(c) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{34} } } ) = P(Z > 2.54) = 1 - P(Z \leq 2.54)

                                                           = 1 - 0.9945 = <u>0.0055</u>

The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 ​minutes, then we conclude that the population mean must be more than 72​, since the probability is so low.

6 0
3 years ago
Jason has 9 Orange balloons he gave Mike 8 of the balloons how many Orange balloons does he now have ​
murzikaleks [220]

Answer: he has 1 orange balloon

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
...........................
Brrunno [24]

Answer:

.. .. .. .. .. .... . ..

Step-by-step explanation:

... .... .. .. ..

8 0
3 years ago
Find mystery number
In-s [12.5K]

The number is less than 190,000 and greater than 180,000

5,000 more than the number has 187 thousands

2000 less than the number has 4 hundreds


180, 000 <n> 190, 000

180,000 + 2000 = n = 5000-187, 000

182,000 = n = 182, 000  

Or,  

 180, 000 <n> 190,000                    

 n + 5000= 187, 000                           n – 2000 = 180, 000

 n= 187, 000- 5000                             n = 180, 000 + 2000

 n= 182, 000                                       n= 182, 000

5 0
4 years ago
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