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3 years ago

15

I guessed “doesn’t matter” but I really want to know why it might be that

1 answer:

chubhunter [2.5K]3 years ago

5 0

I'm sorry, but the spinner that you choose does matter in this case.

For spinner A, there are 5 equal sections with a 1 being in only one of them as well. This gives you a $\frac{1}{5}$ chance of landing on the number 1.

For spinner B, there are 6 equal sections with a 1 being in only one of them. This gives you a $\frac{1}{6}$ chance of landing on the number 1 - a worse chance than if you were to use spinner A.

So for this question, you have a better chance of landing on number 1 if you choose to use spinner A.

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(-4) x 5 + (-6)

Svetllana [295]

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8 0

3 years ago

Measure the lengths of the sides of ∆ABC in GeoGebra, and compute the sine and the cosine of ∠A and ∠B. Verify your calculations

marusya05 [52]

Answer:

$Sin \angle A =0.80$

$Cos \angle A=0.60$

$Sin \angle B =0.60$

$Cos \angle B=0.80$

Step-by-step explanation:

Given

I will answer this question using the attached triangle

Solving (a): Sine and Cosine A

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle A =\frac{BC}{BA}$

Substitute values for BC and BA

$Sin \angle A =\frac{8cm}{10cm}$

$Sin \angle A =\frac{8}{10}$

$Sin \angle A =0.80$

$Cos \angle A=\frac{AC}{BA}$

Substitute values for AC and BA

$Cos \angle A=\frac{6cm}{10cm}$

$Cos \angle A=\frac{6}{10}$

$Cos \angle A=0.60$

Solving (b): Sine and Cosine B

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle B =\frac{AC}{BA}$

Substitute values for AC and BA

$Sin \angle B =\frac{6cm}{10cm}$

$Sin \angle B =\frac{6}{10}$

$Sin \angle B =0.60$

$Cos \angle B=\frac{BC}{BA}$

Substitute values for BC and BA

$Cos \angle B=\frac{8cm}{10cm}$

$Cos \angle B=\frac{8}{10}$

$Cos \angle B=0.80$

Using a calculator:

$A = 53^{\circ}$

So:

$Sin(53^{\circ}) =0.7986$

$Sin(53^{\circ}) =0.80$ -- approximated

$Cos(53^{\circ}) = 0.6018$

$Cos(53^{\circ}) = 0.60$ -- approximated

$B = 37^{\circ}$

So:

$Sin(37^{\circ}) = 0.6018$

$Sin(37^{\circ}) = 0.60$ --- approximated

$Cos(37^{\circ}) = 0.7986$

$Cos(37^{\circ}) = 0.80$ --- approximated

8 0

3 years ago

Read 2 more answers

3.24 x 10-4<br> 8.1 x 10-7

defon

28.4

Explanation

(Multiply the numbers) 3.24 x 10= 32.4
(Subtract the numbers) 32.4 - 4= 28.4

7 0

3 years ago

98 POINTS!!!!! PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Maslowich

Hello!

I believe the answer is C.

I hope it helps!

3 0

3 years ago

Read 2 more answers

\frac{h^{2}-1 }{h+3} when h=5<br><br>Plz answer

Norma-Jean [14]

Answer:

3

Step-by-step explanation:

$\frac{h^{2}-1 }{h+3}$

$\frac{5^{2} -1}{5+3}$

$\frac{25-1}{8}$

$\frac{24}{8}$

3

4 0

3 years ago