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Korvikt [17]
3 years ago
7

The nutritional information on Xuan’s container of crackers said that it had 162 calories. If there were 24 crackers in the cont

ainer, which best describes the number of calories in each cracker?
Mathematics
1 answer:
Colt1911 [192]3 years ago
4 0

Answer:

B. approximately 7 calories

Step-by-step explanation:

got it right, hope you do as well

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What is the solution to the equation 6t = 114
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What is the 100th term if the sequence? 5 ,1 ,-3,-7
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and we also know the first term is 5, ok... then

</span>\bf n^{th}\textit{ term of an arithmetic sequence}\\\\&#10;a_n=a_1+(n-1)d\qquad &#10;\begin{cases}&#10;n=n^{th}\ term\\&#10;a_1=\textit{first term's value}\\&#10;d=\textit{common difference}\\&#10;----------\\&#10;d=-4\\&#10;a_1=5\\&#10;n=100&#10;\end{cases}&#10;\\\\\\&#10;a_{100}=5+(100-1)(-4)\implies a_{100}=5+(99)(-4)&#10;\\\\\\&#10;a_{100}=5-396\implies a_{100}=-391<span>
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3 0
3 years ago
A university researcher wants to estimate the mean number of novels that seniors read during their time in college. An exit surv
lys-0071 [83]

Answer:

Population mean = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

Step-by-step explanation:

Given - A university researcher wants to estimate the mean number

            of  novels that seniors read during their time in college. An exit

            survey was conducted with a random sample of 9 seniors. The

            sample mean was 7 novels with standard deviation 2.29 novels.

To find - Assuming that all conditions for conducting inference have

              been met, which of the following is a 94.645% confidence

              interval for the population mean number of novels read by

              all seniors?

Proof -

Given that,

Mean ,x⁻ = 7

Standard deviation, s = 2.29

Size, n = 9

Now,

Degrees of freedom = df

                                = n - 1

                                = 9 - 1

                                = 8

⇒Degrees of freedom = 8

Now,

At 94.645% confidence level

α = 1 - 94.645%

   =1 - 0.94645

  =0.05355 ≈ 0.05

⇒α = 0.5

Now,

\frac{\alpha}{2} = \frac{0.05}{2}

  = 0.025

Then,

t_{\frac{\alpha}{2}, df }  = 2.306

∴ we get

Population mean = x⁻ ± t_{\frac{\alpha}{2}, df } ×\frac{s}{\sqrt{n} }

                           = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

⇒Population mean = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

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3 years ago
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dezoksy [38]

Answer:

\huge\boxed{v=5}

Step-by-step explanation:

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3 years ago
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