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3 years ago

PLEASE HELP 40 POINTS

Mathematics

2 answers:

Irina18 [472]3 years ago

6 0

Answer: A. 171 degrees B. 79 degrees

Step-by-step explanation: A circles total amount of degrees is 360. You can add the amounts of the other angles and subtract the sum from 360. This will give you the answers to these problems and others like it. :)

Zarrin [17]3 years ago

4 0

X for (a) is 81 and x for (b) is 79. All the measure angles should equal 360

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Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 64% of vehi

ira [324]

Answer:

a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles carry just one person.

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

$P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291$

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

$\mu = np = 92(0.64) = 58.88$

$\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60$

We have to evaluate the probability that more than 47 cars carry just one person.

$P(x \geq 47)$

After continuity correction, we will evaluate

$P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)$

$= 1 - P(z \leq -2.6913)$

Calculation the value from standard normal z table, we have,

$P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%$

0.996 is the probability that more than half out of 92 vehicles carry just one person.

5 0

2 years ago

The x-intercept, or zero, of function g is x = . Function g is over the interval [-5, 5]. As the value of x approaches positive

Aleksandr-060686 [28]

Answer:

boom box second one one my bad just need points my g

Step-by-step explanation:

7 0

2 years ago

Suppose you want to build a fish tank in the shape of a right rectangular box with square base and no top which will hold 6 cubi

AlekseyPX

You must develop a cost function C(x) and then minimize its value.
How much dwill the glass cost? It's $1 per sq ft, and the total area of the glass is 4(xh), where x is the length of one side of the base and h is the height of the tank. The area of the metal bottom is x^2, which we must multiply by $1.50 per sq ft.

This cost function will look like this: C(x) = 4($1/ft^2)xh + ($1.50/ft^2)x^2

but we know that (x^2)h= 6 cu ft, or h = (6 cu ft) / (x^2). Subst. this last result into the C(x) equation, immediately above:

C(x) = 4($1/ft^2)x[6 ft^3 / x^2] + ($1.50/ft^2)x^2

Let's focus on the numerical values and ditch the units of measurement for now:

C(x) = 4x(4/x^2) + 1.50x^2, or

C(x) = 16/x + 1.5x^2

Differentiate this with respect to x:

C '(x) = -16 / x^2 + 3 x

Set this equal to 0 and solve for x: -16/x^2 = -3x, or 16 = 3x^3

Then x^3 = 16/3, and x = 5 1/3 ft. We already have the formula

(x^2)h= 6 cu ft, so if x = 5 1/3, or 16/3, then (16/3)^2 h = 6, or

h = 6 / [16/3]^2.

h = 6 (9/256) = 0.21 ft. While possible, this h = 0.21 ft seems quite unlikely.

Please work through this problem yourself, making sure you understand each step. If questions arise, or if you find an error in my approach, please let me know.

Once again:
1. Write a formula for the total cost of the material used: 4 sides of dimensions xh each, plus 1 bottom, of dimensions x^2. Include the unit prices: $1 per square foot for the sides and $1.50 per square foot for the bottom.
2. Differentiate C(x) with respect to x.
3. Set C '(x) = 0 and solve for the critical value(s).
4. Calculate h from your value for x.
5. Write the dimensions of the tank: bottom: x^2; height: h

4 0

3 years ago

The following question involves a standard deck of 52 playing cards. In such a deck of cards there are four suits of 13 cards ea

olasank [31]

Answer:

(a)No. The probability of drawing a specific second card depends on the identity of the first card.

(b)4/663

(d) 8/663

Step-by-step explanation:

(a)The events are not independent because we are drawing cards without replacement and the probability of drawing a specific second card depends on the identity of the first card.

(b) P(ace on 1st card and jack on 2nd).

$P$(Ace on 1st card) =\dfrac{4}{52}\\ P$(Jack on 2nd card)=\dfrac{4}{51}\\\\$Therefore:\\P(ace on 1st card and jack on 2nd) =\dfrac{4}{52}\times \dfrac{4}{51}\\=\dfrac{4}{663}$

(c)P(jack on 1st card and ace on 2nd)

$P$(Jack on 1st card) =\dfrac{4}{52}\\ P$(Ace on 2nd card)=\dfrac{4}{51}\\\\$Therefore:\\P(jack on 1st card and ace on 2nd) =\dfrac{4}{52}\times \dfrac{4}{51}\\=\dfrac{4}{663}$

(d)Probability of drawing an ace and a jack in either order.

We can either draw an ace first, jack second or jack first, ace second.

Therefore:

P(drawing an ace and a jack in either order) =P(AJ)+(JA)

From parts (b) and (c) above:

$P$(jack on 1st card and ace on 2nd) =\dfrac{4}{663}\\P$(ace on 1st card and jack on 2nd) =\dfrac{4}{663}\\$Therefore:\\P(drawing an ace and a jack in either order)=\dfrac{4}{663}+\dfrac{4}{663}\\=\dfrac{8}{663}$

6 0

3 years ago

Suppose a simple random sample of size 50 is selected from a population with ? = 10. Find the value of the standard error of the

viktelen [127]

Answer:

The population size is N = 50,000

Step-by-step explanation:

8 0

2 years ago