Find the distance between the two points? (1,2),(7,6)

A factory produces 6 times as many loaves of white bread as wholemeal bread in a day. If the factory produces 2640 loaves of whi

kondor19780726 [428]

The number of loaves of white bread more than whole meal bread that are produced in a day are 1760.

We are given that:

A factory produces 6 times as many loaves of white bread as whole meal bread in a day that is:

Let the number of whole meal bread be a variable say x.

So, the number of loaves of white bread = 6 x.

Now, we are given that:

number of loaves of white bread = 2640

6 x = 2640

x = 2640 / 6

x = 880

The number of loaves of white bread more than whole meal bread that are produced in a day = 2640 - 880 = 1760

Therefore, we get that the number of loaves of white bread more than whole meal bread that are produced in a day are 1760.

Learn more about variable here:

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8 0

2 years ago

1) 4x-7y=3 x-7y=-15 2) 9x+8y=15 9x+8y=30

spin [16.1K]

1) x=6, y=3
2) no solution

4 0

3 years ago

Is the height of a person and the person's age linear?Why or why not?

abruzzese [7]

Its not linear because growth rates vary with age, gender also plays a role in the change

6 0

3 years ago

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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

3 × ? = 24 tenths (random words i have to write to ask my question just ignore pls )

antiseptic1488 [7]

Answer: 0.08
explanation: 0.24/3=0.08
3*0.08=0.24

8 0

3 years ago

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