All categories

3 years ago

Please answer 18-20

SAT

1 answer:

Rina8888 [55]3 years ago

8 0

Answer:

-2

Explanation:

u borrow 20 and pay back with 18. What would be left in the credit is 2. So it is therefore -2

You might be interested in

Consider a set of cards that has four cards labeled 1, 2, 3, and 4. Suppose you pick two cards, without replacement, to obtain t

Phantasy [73]

Answer:

$\begin{array}{cccccc}{Mean} & {1.5} & {2} & {2.5} & {3} & {3.5}\ \\ {Probability} & {\frac{1}{6}} & {\frac{1}{6}} & {\frac{1}{3}} & {\frac{1}{6}} & {\frac{1}{6}}\ \end{array}$

Explanation:

Given

$Cards = \{1,2,3,4\}$

Required

The sampling distribution

The possible selection of 2 cards without replacement is as follows:

$S = \{(1,2) (1,3) (1,4) (2,1) (2,3) (2,4) (3,1) (3,2) (3,4) (4,1) (4,2) (4,3)\}$

Calculate the mean

$\begin{array}{cccccccccccc}{Selection} & {(1,2)} & {(1,3)} & {(1,4)} & {(2,1)} & {(2,3)} & {(2,4)}& {(3,1)} & {(3,2)} & {(3,4)} & {(4,1)} & {(4,2)} & {(4,3)} \ \\ {Mean} & {1.5} & {2} & {2.5} & {1.5} & {2.5} & {3} & {2} & {2.5} & {3.5} & {2.5} & {3} & {3.5}\ \end{array}$

List out the mean and the respective frequency

$1.5 \to 2$

$2 \to 2$

$2.5 \to 4$

$3 \to 2$

$3.5\to 2$

$Total \to 12$

Calculate the probability of each mean

$P(1.5) \to \frac{2}{12} \to \frac{1}{6}\\$

$P(2) \to \frac{2}{12} \to \frac{1}{6}\\$

$P(2.5) \to \frac{4}{12} \to \frac{1}{3}\\$

$P(3) \to \frac{2}{12} \to \frac{1}{6}\\$

$P(3.5) \to \frac{2}{12} \to \frac{1}{6}$

So, the table of sampling distribution is:

$\begin{array}{cccccc}{Mean} & {1.5} & {2} & {2.5} & {3} & {3.5}\ \\ {Probability} & {\frac{1}{6}} & {\frac{1}{6}} & {\frac{1}{3}} & {\frac{1}{6}} & {\frac{1}{6}}\ \end{array}$

3 0

3 years ago

Jane is saving to buy a cell phone. She is given a $100.00 gift to start and saves $35 a month from her allowance. So after 1 mo

Bezzdna [24]

Answer:

Yes

Explanation:

becuase she saves the same amount money every month

it's linear

3 0

3 years ago

Read 2 more answers