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3 years ago

Someone help with this plllzz fast

Mathematics

1 answer:

kupik [55]3 years ago

3 0

Since you have the square root of x, you would need to square it to reverse the square root, and just have x. Then, you would have to square the 4 on the otherside, to get x=16

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A box contains 14 nickels, 15 dimes and 6 pennies. if a coin is picked at random from the box, what is the average value of the

Sidana [21]

Answer:

$0.0646 is the expected value of the draw.

Step-by-step explanation:

1 nickel = 5 cents

1 dime = 10 cents

1 penny = 1 cent

Total number of coins = $14+15+6=35$

Probability that a nickel is selected is = 14/35

Probability that a dime is selected is = 15/35

Probability that a penny is selected is = 6/35

So, the expected value of the draw will be =

$\frac{14}{35}*5+\frac{15}{35}*10+\frac{6}{35}*1$

$\frac{70}{35}+\frac{150}{35}+\frac{6}{35}=\frac{226}{35}$

= 6.457 cents ≈6.46 cents

So, in dollars it is = $\frac{6.46}{100}$ = $0.0646

So, $0.0646 is the expected value of the draw.

5 0

3 years ago

Charra [1.4K]

Answer:

Yes its similar because if you look at all the numbers they are all the same but the shape is just turned around

please consider brainliest <3

4 0

2 years ago

Read 2 more answers

Evaluate the expression where x= 5 y=2

PtichkaEL [24]

4/7 because the equation would be 2(7-5)
over
5+2
so it would be
2(2)
over
7

and then it would be
4
over
7

4 0

3 years ago

PLEASE HELP ASAP EVERYTHING IS DUE TM

Ronch [10]

The first three go in the first three boxes. Hoped I helped! Mark me brainly if can

4 0

3 years ago

Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL

Mice21 [21]

Answer:

$KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47$

Step-by-step explanation:

Given:

KL ║ NM ,

LM = 45

m∠M = 50°

KN ⊥ NM

NL ⊥ LM

Find: KN and KL

1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

LM = 45

m∠M = 50°

So,

$\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}$

Also

$m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ}$ (angles LNM and M are complementary).

2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

$NL=45\tan 50^{\circ}$

$m\angle KLN=m\angle LNM=40^{\circ}$ (alternate interior angles)

$m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ}$ (angles KNL and KLN are complementary).

So,

$\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08$

and

$\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47$

3 0

3 years ago

Read 2 more answers