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2 years ago

8

From a random sample of 92 male students at Hope College, 8 were left-handed. Using the Theory Based Inference applet, determine

a 99% confidence interval for the proportion of all male students at Hope College that are left -handed.

1 answer:

Bumek [7]2 years ago

3 0

Answer:

mkkkmnsine kgsk jtb jgjdb Geeta kl John ku

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Determine the coordinates of the vertices of the rectangle to compite the area of the rectangle using the distance formula.

Veseljchak [2.6K]

Answer:

Step-by-step explanation:

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3 years ago

What is 1/3 of 3/4?

ad-work [718]

I really don't understand what your saying but.....
1/3 = 0.3333333...... and 3/4 = 0.75 = 75%
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3 years ago

The ammunition storage room has 10 feet between the floor and the ceiling. Each box of ammunition is 10 feet tall. Each crate of

KIM [24]

Answer:

the maximum number of crates that can be stacked between the floor and ceiling

$\frac{height \hspace{0.1cm} of \hspace{0.1cm} ammunition \hspace{0.1cm} box}{height \hspace{0.1cm} of \hspace{0.1cm} each \hspace{0.1cm} crate} = \frac{10 \hspace{0.1cm} feet}{12 \hspace{0.1cm} inches} = \frac{10 \times 12 \hspace{0.1cm} inches }{12 \hspace{0.1cm} inches} = 10 \hspace{0.1cm} crates$

where 1 foot = 12 inches. SO the answer is that a maximum of 10 crates can be stacked from floor to ceiling.

Step-by-step explanation:

i) the maximum number of crates that can be stacked between the floor and ceiling

$\frac{height \hspace{0.1cm} of \hspace{0.1cm} ammunition \hspace{0.1cm} box}{height \hspace{0.1cm} of \hspace{0.1cm} each \hspace{0.1cm} crate} = \frac{10 \hspace{0.1cm} feet}{12 \hspace{0.1cm} inches} = \frac{10 \times 12 \hspace{0.1cm} inches }{12 \hspace{0.1cm} inches} = 10 \hspace{0.1cm} crates$

where 1 foot = 12 inches

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3 years ago

Does someone know how to do this

Sveta_85 [38]

Answer:

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Step-by-step explanation:

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8 0

3 years ago

in a recent semester at a local university 520 students enrolled in both general chemistry and calculus 1. Of these students 88

IceJOKER [234]

Let Ch and C denote the events of a student receiving an A in <u>ch</u>emistry or <u>c</u>alculus, respectively. We're given that

P(Ch) = 88/520

P(C) = 76/520

P(Ch and C) = 31/520

and we want to find P(Ch or C).

Using the inclusion/exclusion principle, we have

P(Ch or C) = P(Ch) + P(C) - P(Ch and C)

P(Ch or C) = 88/520 + 76/520 - 31/520

P(Ch or C) = 133/520

6 0

3 years ago