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3 years ago

Help ! Taking the test right now

Mathematics

1 answer:

Zarrin [17]3 years ago

5 0

The answer is the third one

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PLEASE ANSWER!!! <br> You don’t have to do number 25 *****

Ilia_Sergeevich [38]

Answer:

see explanation

Step-by-step explanation:

The directed lines have the form < x, y >

With the usual notation

right in x- direction → +

left in x- direction → -

up in y- direction → +

down in y- direction → -

EF = < 4, 1 >

FG = < 4, - 2 >

GH = < - 4, - 2 >

EH = < 4, - 3 >

8 0

2 years ago

The model represents an equation. What value of x makes the equation true?

Anuta_ua [19.1K]

Answer:

I think i wanna say one of the fractions

Step-by-step explanation:

5 0

2 years ago

Dr. Nefario put $5000 in the bank and left it there for 6 months. What simple interest rate would he need to have in order to ha

Sonja [21]

Answer: The interest rate is .166%

Step-by-step explanation:

So what I did was divide $50(Amount at the end) by 6 (the number of months). Which would give me the amount of money he made per month ($8.333...). So then $8.333.../$5000(Total $) is .166...% (Interest rate)

7 0

2 years ago

Read 2 more answers

The table shows the relationship between a, the area of a rectangle, and h, its height, when the base remains constant.

olganol [36]

Answer:

a = 4h.

Step-by-step explanation:

Area (a) = base (b) * height (h).

When h = 2, a = 8:

a = bh

6 = 2b

b = 4

The result 4 is the same for h = 5, 7 and 12.

So the required equation is

a = 4h.

4 0

3 years ago

Three cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is an ace,

MrRissso [65]

Answer:

$4.82\cdot 10^{-4}$

Step-by-step explanation:

In a deck of cart, we have:

a = 4 (aces)

t = 4 (three)

j = 4 (jacks)

And the total number of cards in the deck is

n = 52

So, the probability of drawing an ace as first cart is:

$p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769$

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

$n-1=51$

Therefore, the probability of drawing a three at the 2nd draw is

$p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784$

Then, at the third draw, the previous 2 cards are not replaced, so there are now

$n-2=50$

cards in the deck. So, the probability of drawing a jack is

$p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08$

Therefore, the total probability of drawing an ace, a three and then a jack is:

$p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}$

4 0

3 years ago