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1 year ago

Find the measure of the exterior 1. A. 11° OB. 127° O C. 85° O D. 95°

Mathematics

1 answer:

Marizza181 [45]1 year ago

7 0

Answer: D) 95°

Step-by-Step Solution:

We know that the Exterior Angle of a Triangle = Sum of Opposite Interior Angles of the Triangle.

Therefore,

∠1 = 53° + 42°
=> ∠1 = 95°

Hence, ∠1 = 95°

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the nth term of a sequence is given by an²+bn where a and b are integers the 2nd term of the sequence is 2 4th term of the seque

stealth61 [152]

Answer:

Step-by-step explanation:

Tn = an²+bn

T2 = a2²+2b

2 = 4a +2b

Multiply all through by 2

4 = 8a +4b

4b = 4-8a

T4 = a4² + 4b

20 = 16a +4b

4b = 20-16a

4-8a = 20-16a

16a -8a = 20-4

8a = 16

a = 16/8 = 2

4b = 4-8a = 4-8(2) = 4-16 = -12

b = -12/4 = -3

T6 = a6² + 6b

T6 = 2(36) + 6(-3) = 72 -18 = 54

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2 years ago

Read 2 more answers

Please answer this it would be very helpful if you did.

Dmitriy789 [7]

Answer:

See below ↓↓↓

Step-by-step explanation:

Probability of getting heads

Outcome (head) = 1
Total outcomes = 2
P (head) = 1/2

Probability of getting an odd number

Outcome (odd) = {1, 3, 5} = 3
Total outcomes = 6
P (odd) = 3/6 = 1/2

The outcomes are equally likely because the probabiity of each outcome is the same.

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2 years ago

Which number has a repeating decimal form?

blondinia [14]

Answer:

D. $\frac{2}{6}$

Step-by-step explanation:

2÷ 6 = 0.33333

As you can see 3 is repeating.

Therefore $\frac{2}{6}$ as repeating decimal form.

3 0

2 years ago

A Simple random sample of 100 8th graders at a large suburban middle school indicated that 84% of them are involved with some ty

Setler [38]

Answer: a) (0.755, 0.925)

Step-by-step explanation:

Let p be the population proportion of 8th graders are involved with some type of after school activity.

As per given , we have

n= 100

sample proportion: $\hat{p}=0.84$

Significance level : $\alpha= 1-0.98=0.02$

Critical z-value : $z_{\alpha/2}=2.33$ (using z-value table)

Then, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity will be :-

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

i.e. $0.84\pm (2.33)\sqrt{\dfrac{0.84(1-0.84)}{100}}$

i.e. $\approx0.84\pm 0.085$

i.e. $(0.84- 0.085,\ 0.84+ 0.085)=(0.755,\ 0.925)$

Hence, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity : a) (0.755, 0.925)

3 0

3 years ago

Let f(x)equals=33xminus−1, h(x)equals=startfraction 7 over x plus 5 endfraction 7 x+5 . find (hcircle◦f)(66).

Ronch [10]

$\bf \begin{cases}
f(x)=3x-1\\
h(x)=\cfrac{7}{x+5}\\\\
(h\circ f)(x)=h(~~f(x)~~)
\end{cases}
\\\\\\
f(66)=3(66)-1\implies \boxed{f(66)=197}
\\\\\\
\stackrel{h(~~f(x)~~)}{h(~~f(66)~~)}\implies \stackrel{h(~~f(66)~~)}{h\left(~~\boxed{197}~~ \right)}=\cfrac{7}{\boxed{197}+5}\implies h\left(~~\boxed{197}~~ \right)=\cfrac{7}{202}$

7 0

3 years ago