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lapo4ka [179]
3 years ago
6

30.48 kg of beef was placed into 24 packages of equal weight. What is the weight of one package of beef?

Mathematics
2 answers:
Ad libitum [116K]3 years ago
6 0

Answer:

1 packet weight =(30.48 kg)/24

=1.27 kg

mario62 [17]3 years ago
5 0

Answer:

the weight of one package of beef is 1.27 kg

Step-by-step explanation:

mass of beef = 30.48 kg

no of packet = 24

to find out

What is the weight of one package of beef

solution

we will apply here formula for find 1 packet beef that is

1 packet weight = \frac{total mass}{total packet }   ...................1

put here value in equation 1

1 packet weight = \frac{30.48}{24}  

1 packet weight = 1.27

so the weight of one package of beef is 1.27 kg

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Step-by-step explanation:

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Mako runs 10 miles in 2 hours. If he continues at this rate, how many miles
Vesnalui [34]

Answer:

Answer: 25 miles

Step-by-step explanation:

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3 years ago
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Solve the system of equations using the linear combination method {5p-3q=-39 -2p-3q=3
VikaD [51]

Answer:

p=-6

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Step-by-step explanation:

5p-3q=-39

-2p-3q=3

Multiply the second equation by -1

2p +3q = -3

Add the first equation and the modified second equation

5p-3q=-39

2p +3q = -3

---------------------

7p = -42

Divide by 7

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Add 12 to each side

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Test the claim that the mean GPA of night students is larger than 3.1 at the .10 significance level. The null and alternative hy
Alborosie

Answer:

H 0 : μ = 3.1    H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66

df = n-1=75-1 =74

The p value is:

p_v = P(t_{74} >8.66) =3.65x10^{-13}

The significance level is: 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

Step-by-step explanation:

For this case we want to test the claim that mean GPA of night students is larger than 3.1 at the .10 significance level. The claim needs to be on the alternative hypothesis so then we have the following system of hypothesis:

H 0 : μ = 3.1    H 1 : μ > 3.1

And this test is defined a a right tailed test since the symbol for the alternative hypothesis H1 is >.

We have the following info given:

\bar X = 3.13 , s =0.03 , n =75

The statistic to check the hypothesis is given by:

t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

Replacing the info given we got:

t = \frac{3.13-3.1}{\frac{0.03}{\sqrt{75}}}= 8.66

The degrees of freedom are given by:

df = n-1=75-1 =74

The p value since is a right tailed ted is given by:

p_v = P(t_{74} >8.66) =3.65x10^{-13}

The significance level is 0.1 or 10%

And since the p value is very low compared to the significance level we have enough evidence to:

Reject the null hypothesis

6 0
3 years ago
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