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Viktor [21]
2 years ago
6

The average temperature for four days days was 22 degrees what must the temperature be on the fifth day in order to make 23 degr

ees the average temperature for the five days
Mathematics
1 answer:
galina1969 [7]2 years ago
5 0

Answer:

27 degrees.

Step-by-step explanation:

If the average is to be 23 degrees over 5 days then the total of the temperatures will be 5*23 = 115 degrees.

The total over the first 4 days = 22*4 = 88 degrees, so the temperature on the fifth day has to be 115 - 88 = 27 degrees.

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Mandy can buy 4 containers of yogurt and 3 boxes of crackers for $9.55. She can buy 2 containers of yogurt and 2 boxes of cracke
Anna11 [10]

Answer:

The cost of a box of crackers is $2.25.

Step-by-step explanation:

Let the cost of a containers of yogurt is $x and cost of a box of crackers is $y.

It is given that Mandy can buy 4 containers of yogurt and 3 boxes of crackers for $9.55.

4x+3y=9.55              .... (1)

She can buy 2 containers of yogurt and 2 boxes of crackers for $5.90.

2x+2y=5.90              .... (2)

On solving (1) and (2), using graphing calculator, we get

x=0.7

y=2.25

Therefore the cost of a box of crackers is $2.25.

6 0
2 years ago
Read 2 more answers
A researcher measures IQ and weight for a group of college students. What kind of correlation is likely to be obtained for these
Sunny_sXe [5.5K]

Answer:

Option b

Step-by-step explanation:

Given that a  researcher measures IQ and weight for a group of college students.

In general, we think that the weight has nothing to do with IQ of a person and hence not correlated.

But if we go deep, we find that after a certain weight, the person becomes lazy and inactive with a chance to have reduced IQ

Weight gain causes also health problems including less activity of both brain and body and hence there is a chance for less IQ

So we find that as weight increases iq decreases and when weight decreases, IQ increases.

Thus we can say that there is a negative correlation but not necessarily near to one.

Hence option b is right

8 0
2 years ago
A study of consumer smoking habits includes 177 people in the 18-22 age bracket ( 48 of whom smoke), 146 people in the 23-30 age
svp [43]

Answer:

The probability of getting someone who is age 18-22 or does not smoke is 0.854 ....

Step-by-step explanation:

Through the given statements we have to find the probability of getting someone who is age 18-22 or does not smoke.

Age 18-22 has 177 people. 48 of whom smoke.

People who does not smoke =(177-48) + (146-31) +(81-28)

People who does not smoke=129+115+53 = 297

People with age (18-22) who does not smoke = 129

P(18-22 or does not smoke) = (177+297-129)/(177+146+81)

P(18-22 or does not smoke) = 345/404

P(18-22 or does not smoke) = 0.854

Thus the probability of getting someone who is age 18-22 or does not smoke is 0.854 ....

3 0
2 years ago
Read 2 more answers
A sample of tritium-3 decayed to 94.5% of its original amount after a year.
Zolol [24]
(a) If y(t) is the mass after t days and y(0) = A then y(t) = Ae^{kt}.

y(1) = Ae^k = 0.945A \implies e^k = 0.945 \implies k = \ln 0.945

\text{Then } Ae^{(\ln 0.945)t} = \frac{1}{2}A\ \Leftrightarrow\ \ln e^{(\ln 0.945)t} = \ln \frac{1}{2}\ \Leftrightarrow\  ( \ln 0.945) t = \ln \frac{1}{2} \Leftrightarrow \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 12.25 \text{ years}

(b)
Ae^{(\ln 0.945) t} = 0.20 A \ \Leftrightarrow\ (\ln 0.945) t \ln \frac{1}{5}\ \implies \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 28.45\text{ years}
3 0
2 years ago
Searchers are planning a study to estimate the impact on crop yield when no-till is used in combination with residue retention a
VladimirAG [237]

Answer:

n=77

Step-by-step explanation:

Notation and definitions

\hat p=0.025 or 2.5%, estimated proportion for decline in crop yield when no-till is used with residue retention and crop rotation

p true population proportion for decline in crop yield when no-till is used with residue retention and crop rotation

n (variable of interest) represent the sample size required

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.035 or 3.5% and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

And replacing into equation (b) the values from part a we got:

n=\frac{0.025(1-0.025)}{(\frac{0.035}{1.96})^2}=76.44  

And rounded up we have that n=77

5 0
2 years ago
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