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3 years ago

Lines a and b in the diagram are parallel.

Mathematics

2 answers:

Romashka [77]3 years ago

6 0

Answer:

The answer is c cause I took the quiz

Step-by-step explanation:

uysha [10]3 years ago

3 0

<h2>Answer:</h2>

C.) m∠C = 45°

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Find all solutions to the equation in the interval [0, 2pi). COS X = sin 2x

Lera25 [3.4K]

Answer:

Step-by-step explanation:

cos x=sin 2x

sin 2x-cos x=0

2 sin x cos x-cos x=0

cos x(2 sin x-1)=0

either cos x=0

$x=\frac{\pi }{2},\frac{3\pi }{2}$

2 sin x-1=0

$sin x=\frac{1}{2} =sin (\frac{\pi }{6}),sin(\pi -\frac{\pi }{6} )\\x=\frac{\pi }{6} ,\frac{5 \pi }{6}$

6 0

3 years ago

During the 7th examination of the Offspring cohort in the Framingham Heart Study, there were 1219 participants being treated for

AlexFokin [52]

Answer:

95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

Step-by-step explanation:

We are given that during the 7th examination of the Offspring cohort in the Framing ham Heart Study, there were 1219 participants being treated for hypertension and 2,313 who were not on treatment.

The sample proportion is : $\hat p$ = x/n = 1219/3532 = 0.345

Firstly, the pivotal quantity for 95% confidence interval for the proportion of the population is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion = 0.345

n = sample of participants = 3532

p = population proportion

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p= [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.345-1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ , $0.345+1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ ]

= [0.3293 , 0.3607]

Hence, 95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

6 0

3 years ago

Give the first ten terms of the following sequences. You can assume that the sequences start with an index of 1. Logs are to bas

Vitek1552 [10]

Answer:

a1 = log(1) = 0 (2⁰ = 1)

a2 = log(2) = 1 (2¹ = 2)

a3 = log(3) = ln(3)/ln(2) = 1.098/0.693 = 1.5849

a4 = log(4) = 2 (2² = 4)

a5 = log(5) = ln(5)/ln(2) = 1.610/0.693 = 2.322

a6 = log(6) = log(3*2) = log(3)+log(2) = 1.5849+1 = 2.5849 (here I use the property log(a*b) = log(a)+log(b)

a7 = log(7) = ln(7)/ln(2) = 1.9459/0.6932 = 2.807

a8 = log(8) = 3 (2³ = 8)

a9 = log(9) = log(3²) = 2*log(3) = 2*1.5849 = 3.1699 (I use the property log(a^k) = k*log(a) )

a10 = log(10) = log(2*5) = log(2)+log(5) = 1+ 2.322= 3.322

b) I can take the results of log n we previously computed above to calculate 2^log(n), however the idea of this exercise is to learn about the definition of log_2:

log(x) is the number L such that 2^L = x. Therefore 2^log(n) = n if we take the log in base 2. This means that

a1 = 1

a2 = 2

a3 = 3

a4 = 4

a5 = 5

a6 = 6

a7 = 7

a8 = 8

a9 = 9

a10 = 10

I hope this works for you!!

3 0

3 years ago

The dimensions of a rectangle are 3 cm by 5

aleksley [76]

9 or 12 I think Possibly 18

6 0

4 years ago

Read 2 more answers

Seven less than the product of a number and 10 is 50

Veseljchak [2.6K]

product of a number and 10: 10n
Seven less than the product of a number and 10: 10n - 7
so
Seven less than the product of a number and 10 is 50:
10n - 7 = 50
----------------------------
if you need to find the number n then
10n - 7 = 50
10n = 57
n = 57/10
n = 5.7

5 0

3 years ago