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2 years ago

5

144 in the ratio 3:4

1 answer:

Nataly_w [17]2 years ago

5 0

Answer: 132

Step-by-step explanation:

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HELPP, Lines parallel, Perpendicular, Neither? URGENT HELPPP.

VLD [36.1K]

Answer:

perpendicular I believe

Step-by-step explanation:

I'd plot the points and make a line using a ruler to check though.

8 0

2 years ago

Can y’all help me with it pls

Marta_Voda [28]

Answer:

1. D

2. C

3. C, A F

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

If A and B are independent events with P(A) = .5 and P(B) = .2, find the following:a) P(A U B)b) P(A^c ? B^c)c) P(A^c U B^c)**No

Effectus [21]

If A and B are independent, then $P(A\cap B)=P(A)P(B)$ .

a.

$P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)$

$P(A\cup B)=0.5+0.2-0.5\cdot0.2$

$\boxed{P(A\cup B)=0.6}$

b. I'm guessing the ? is supposed to stand for intersection. We can use DeMorgan's law for complements here:

$P(A^c\cap B^c)=P(A\cup B)^c=1-P(A\cup B)$

$P(A^c\cap B^c)=1-0.6$

$\boxed{P(A^c\cap B^c)=0.4}$

c. DeMorgan's law can be used here too:

$P(A^c\cup B^c)=P(A\cap B)^c=1-P(A\cap B)=1-P(A)P(B)$

$P(A^c\cup B^c)=1-0.5\cdot0.2$

$\boxed{P(A^c\cup B^c)=0.9}$

4 0

2 years ago

What is happening to this graph when the x-values are between 4 and 8?

Vilka [71]

Answer:

The correct answer is A. It is constant.

Step-by-step explanation:

When the x-values are between 4 and 8, this graph is constant. This can be seen because the graph is a horizontal line, meaning it has zero slope. Since the slope is zero, it is neither increasing nor decreasing, and is known as constant.

Hope this helps!

6 0

3 years ago

In the △ABC, the height AN = 24 in, BN = 18 in, AC = 40 in. Find AB and BC. Consider all possible cases. Case 1 : N∈BC, AB = __,

aivan3 [116]

Answer:

Case 1:

$AB = 30$

$BC = 50$

Case 2:

$AB = 15.9$

$BC = 36.7$

Case 3: Not possible

Step-by-step explanation:

Given

See attachment for illustration of each case

Required

Find AB and BC

Case 1:

Using Pythagoras theorem in ANB, we have:

$AB^2 = AN^2 + BN^2$

This gives:

$AB^2 = 24^2 + 18^2$

$AB^2 = 576 + 324$

$AB^2 = 900$

Take square roots of both sides

$AB = \sqrt{900$

$AB = 30$

To calculate BC, we consider ANC, where:

$AC^2 = AN^2 + NC^2$

$40^2 = 24^2 + NC^2$

$1600 = 576 + NC^2$

Collect like terms

$NC^2 = 1600 - 576$

$NC^2 = 1024$

Take square roots

$NC = \sqrt{1024$

$NC = 32$

So:

$BC = NC + BN$

$BC = 32 + 18$

$BC = 50$

Case 2:

Using Pythagoras theorem in ANB, we have:

$AN^2 = AB^2 + BN^2$

This gives:

$24^2 = AB^2 + 18^2$

$576 = AB^2 + 324$

Collect like terms

$AB^2 = 576 - 324$

$AB^2 = 252$

Take square roots of both sides

$AB = \sqrt{252$

$AB = 15.9$

To calculate BC, we consider ABC, where:

$AC^2 = AB^2 + BC^2$

$40^2 = 252 + BC^2$

$1600 = 252 + BC^2$

Collect like terms

$BC^2 = 1600 - 252$

$BC^2 = 1348$

Take square roots

$BC = \sqrt{1348$

$BC = 36.7$

Case 3:

This is not possible because in ANC

The hypotenuse AN (24) is less than AC (40)

3 0

2 years ago