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LekaFEV [45]
2 years ago
9

Juan wants to start a lemonade stand to earn money.

Mathematics
2 answers:
masha68 [24]2 years ago
7 0

Answer:

whats the question then

topjm [15]2 years ago
7 0

Answer:

1. 13 2. 10

Step-by-step explanation:

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3 years ago
The z-score associated with 95% is 1.96. If the sample mean is 200 and the standard deviation is 30, find the upper limit of the
ziro4ka [17]

Answer:

The upper limit of the 95% confidence interval is:

C.I_u = 200 + (58.8/\sqrt{n})

Step-by-step explanation:

The formula is given as:

C.I = μ ± Z*σ/\sqrt{n}

The upper limit => C.I_u = μ + Z*σ/\sqrt{n}

The lower limit => C.I_l = μ - Z*σ/\sqrt{n}

The sample size (n) is not stated in the question. Hence, we calculate the upper limit with respect to n.

The upper limit => C.I_u = 200 + 1.96*(30/\sqrt{n})

                                    = 200 + (1.96*30)/\sqrt{n}

                                    = 200 + 58.8/\sqrt{n}

5 0
2 years ago
46. Order these numbers from least to greatest:<br> 1/2, 0.4, 1/3, 1/5
e-lub [12.9K]

Answer:

1/5, 1/3, 0.4, 1/2

Step-by-step explanation:

1/5 = 0.2

1/3 = 0.33

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