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FinnZ [79.3K]
3 years ago
15

a bakery sold 31 bagels in the first hour of business and 42 bagels in the second hour. if the baker had 246 bagels to start, ho

w many bagels were left after the second hour?
Mathematics
1 answer:
bogdanovich [222]3 years ago
7 0

Answer:

173 bagels were left after the second hour!!

Step-by-step explanation:

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In one study, it was found that the correlation coefficient between two variables is -0.93.
Marysya12 [62]

the answer is d (i think) because anything aboveboard .90, whether it is positive or negative

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Determine the range of the following graph
marta [7]
The range is the y-values. Here is the answer and explanation :)

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At which point is 731 located on the number line?
ch4aika [34]

Answer:

point S is where 731 is located on the number line

4 0
3 years ago
A book claims that more hockey players are born in January through March than in October through December. The following data sh
astra-53 [7]

Answer:

\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295

Now we can calculate the degrees of freedom for the statistic given by:

df=(categories-1)=4-1=3

And we can calculate the p value given by:

p_v = P(\chi^2_{3} >18.295)=0.00038

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the​ year

Step-by-step explanation:

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference of birthdates distributed throughout the​ year

H1: There is a difference between birthdates distributed throughout the​ year

The level of significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{total}{4}

And replacing we got:

E_{1} =\frac{67+56+30+37}{4}=47.5

And now we can calculate the statistic:

\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295

Now we can calculate the degrees of freedom for the statistic given by:

df=(categories-1)=4-1=3

And we can calculate the p value given by:

p_v = P(\chi^2_{3} >18.295)=0.00038

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the​ year

3 0
4 years ago
What is the answer to this image
tangare [24]

Answer:

EA = \frac{50}{3}

Step-by-step explanation:

Δ CEA and Δ CDB are similar by the AA postulate. Then the ratios of corresponding sides are in proportion , that is

\frac{EA}{DB} = \frac{CE}{CD} ( substitute values )

\frac{EA}{10} = \frac{9+6}{9} = \frac{15}{9} ( cross- multiply )

9 EA = 150 ( divide both sides by 9 )

EA = \frac{150}{9} = \frac{50}{3}

6 0
2 years ago
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