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Nady [450]
3 years ago
5

Urelia made a deposit to her checking account. She had $104.00 in currency; $7.64 in coins; and checks for $83.29, $257.77, $1,3

32.68, and $3,984.05. What was her total deposit?​
Mathematics
1 answer:
pogonyaev3 years ago
3 0
The correct answer is 5769.43
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A jeweler wants to minimize business costs and has found that the average cost in dollars per necklace is given by C(x)=0.5x^2-7
Vika [28.1K]

Answer:

The maximum cost is 41 dollars per necklace.

Step-by-step explanation:

The average cost in dollars/necklace is given by :

C(x)=0.5x^2-7x+65.5  .....(1)

Where

x is the number of necklaces that are created

To maximize the cost, find the value of x by putting dC/dx = 0

\dfrac{d}{dx}(0.5x^2-7x+65.5)=0\\\\x-7=0\\\\x=7

Now put the value of x = 7 in equation (1).

C(x)=0.5(7)^2-7(7)+65.5\\\\=41\ \text{dollars per necklace}

Hence, the maximum cost is 41 dollars per necklace.

8 0
2 years ago
26. Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics
hjlf

The test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

<h3>What are null hypotheses and alternative hypotheses?</h3>

In null hypotheses, there is no relationship between the two phenomenons under the assumption or it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.

Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics course.

Then they are given a post-test once the professor has concluded lecturing on the material.

Pre-test and post-test scores for 4 students in an elementary statistics class are given below.

Then we have

\mu _d = \mu _{post} - \mu _{pre}

Then the null hypotheses and alternative hypotheses will be

H₀: \mu _d = 0

Hₐ: \mu _d > 0

Then the test statistic will be

\rm \overline{x} _d = \dfrac{\Sigma x_d}{n} = \dfrac{15+12+10+1}{4}\\\\\overline{x} _d = 9.5

Then

\rm S_d = 6.02

The test statistic value is given by

\rm t = \dfrac{\overline{x} _d }{\dfrac{S_d}{\sqrtn}} \\\\t = \dfrac{9.5}{\dfrac{6.02}{\sqrt4}}\\\\t = 3.16

Since this is a right-tailed test, so the critical value is given by

\rm t_{n-1}(\alpha ) = t_3 (0.05) = 2.353

Since the test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

Hence, we can conclude that \mu _d > 0 that is test scores have improved.

More about the null hypotheses and alternative hypotheses link is given below.

brainly.com/question/9504281

#SPJ1

7 0
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