Three times the sum of a number and 3 is 3 less than the number times 5. Find the number.

-22 52 -33 71 -44 What is the x-intercept of the line?

swat32

Answer:

30

Step-by-step explanation:

5 0

3 years ago

He drove at Puerto Princesa City to Taytay at an average speed of 40 kph

valkas [14]

Answer:

210km

Step-by-step explanation:

40 x 5=200

40/4=10

10+200=210

distance = speed x time

8 0

3 years ago

Find the sum or difference. a. -121 2 + 41 2 b. -0.35 - (-0.25)

s344n2d4d5 [400]

Answer:

2

Step-by-step explanation:

The reason an infinite sum like 1 + 1/2 + 1/4 + · · · can have a definite value is that one is really looking at the sequence of numbers

1

1 + 1/2 = 3/2

1 + 1/2 + 1/4 = 7/4

1 + 1/2 + 1/4 + 1/8 = 15/8

etc.,

and this sequence of numbers (1, 3/2, 7/4, 15/8, . . . ) is converging to a limit. It is this limit which we call the "value" of the infinite sum.

How do we find this value?

If we assume it exists and just want to find what it is, let's call it S. Now

S = 1 + 1/2 + 1/4 + 1/8 + · · ·

so, if we multiply it by 1/2, we get

(1/2) S = 1/2 + 1/4 + 1/8 + 1/16 + · · ·

Now, if we subtract the second equation from the first, the 1/2, 1/4, 1/8, etc. all cancel, and we get S - (1/2)S = 1 which means S/2 = 1 and so S = 2.

This same technique can be used to find the sum of any "geometric series", that it, a series where each term is some number r times the previous term. If the first term is a, then the series is

S = a + a r + a r^2 + a r^3 + · · ·

so, multiplying both sides by r,

r S = a r + a r^2 + a r^3 + a r^4 + · · ·

and, subtracting the second equation from the first, you get S - r S = a which you can solve to get S = a/(1-r). Your example was the case a = 1, r = 1/2.

In using this technique, we have assumed that the infinite sum exists, then found the value. But we can also use it to tell whether the sum exists or not: if you look at the finite sum

S = a + a r + a r^2 + a r^3 + · · · + a r^n

then multiply by r to get

rS = a r + a r^2 + a r^3 + a r^4 + · · · + a r^(n+1)

and subtract the second from the first, the terms a r, a r^2, . . . , a r^n all cancel and you are left with S - r S = a - a r^(n+1), so

(IMAGE)

As long as |r| < 1, the term r^(n+1) will go to zero as n goes to infinity, so the finite sum S will approach a / (1-r) as n goes to infinity. Thus the value of the infinite sum is a / (1-r), and this also proves that the infinite sum exists, as long as |r| < 1.

In your example, the finite sums were

1 = 2 - 1/1

3/2 = 2 - 1/2

7/4 = 2 - 1/4

15/8 = 2 - 1/8

and so on; the nth finite sum is 2 - 1/2^n. This converges to 2 as n goes to infinity, so 2 is the value of the infinite sum.

8 0

3 years ago

The Salk polio vaccine experiment in 1954 focused on the effectiveness of the vaccine in combating paralytic polio. Because it w

sdas [7]

Answer:

Step-by-step explanation:

Hello!

The variables of interest are:

X₁: Number of cases of polio observed in kids that received the placebo vaccine.

n₁= 201299 total children studied

x₁= 110 cases observed

X₂: Number of cases of polio observed in kids that received the experimental vaccine.

n₂= 200745 total children studied

x₂= 33 cases observed

These two variables have a binomial distribution. The parameters of interest, the ones to compare, are the population proportions: p₁ vs p₂

You have to test if the population proportions of children who contracted polio in both groups are different: p₂ ≠ p₁

a)

H₀: p₂ = p₁

H₁: p₂ ≠ p₁

α: 0.05

$Z= \frac{(p'_2-p'_1)-(p_2-p_1)}{\sqrt{p'[\frac{1}{n_1} +\frac{1}{n_2} ]} }$

Sample proportion placebo p'₁= x₁/n₁= 110/201299= 0.0005

Sample proportion vaccine p'₂= x₂/n₂= 33/200745= 0.0002

Pooled sample proportion p'= (x₁+x₂)/(n₁+n₂)= (110+33)/(201299+200745)= 0.0004

$Z_{H_0}= \frac{(0.0002-0.0005)-0}{\sqrt{0.0004[\frac{1}{201299} +\frac{1}{200745} ]} }= -4.76$

This test is two-tailed, using the critical value approach, you have to determine two critical values:

$Z_{\alpha/2}= Z_{0.025}= -1.96$

$Z_{1-\alpha /2}= Z_{0.975}= 1.96$

Then if $Z_{H_0}$ ≤ -1.96 or if $Z_{H_0}$ ≥ 1.96, the decision is to reject the null hypothesis.

If -1.96 < $Z_{H_0}$ < 1.96, the decision is to not reject the null hypothesis.

⇒ $Z_{H_0}$ = -4.76, the decision is to reject the null hypothesis.

b)

H₀: p₂ = p₁

H₁: p₂ ≠ p₁

α: 0.01

$Z= \frac{(p'_2-p'_1)-(p_2-p_1)}{\sqrt{p'[\frac{1}{n_1} +\frac{1}{n_2} ]} }$

The value of $Z_{H_0}$ = -4.76 doesn't change, since we are working with the same samples.

The only thing that changes alongside with the level of significance is the rejection region:

$Z_{\alpha /2}= Z_{0.005}= -2.576$

$Z_{1-\alpha /2}= Z_{0.995}= 2.576$

Then if $Z_{H_0}$ ≤ -2.576or if $Z_{H_0}$ ≥ 2.576, the decision is to reject the null hypothesis.

If -2.576< $Z_{H_0}$ < 2.576, the decision is to not reject the null hypothesis.

⇒ $Z_{H_0}$ = -4.76, the decision is to reject the null hypothesis.

c)

Remember the level of significance (probability of committing type I error) is the probability of rejecting a true null hypothesis. This means that the smaller this value is, the fewer chances you have of discarding the true null hypothesis. But as you know, you cannot just reduce this value to zero because, the smaller α is, the bigger β (probability of committing type II error) becomes.

Rejecting the null hypothesis using different values of α means that there is a high chance that you reached a correct decision (rejecting a false null hypothesis)

I hope this helps!

8 0

3 years ago

In abc above,what is the length of ad

Nezavi [6.7K]

Answer:

B

Step-by-step explanation:

First calculate BD using sine ratio in Δ BCD and the exact value

sin60° = $\frac{\sqrt{3} }{2}$ , thus