1) graph a 2) graph b 3) graph c 4) graph d

An aircraft carrier left Port 35 traveling north five hours before a container ship. The container ship traveled in the opposite

Molodets [167]

Answer:

Aircraft carrier speed = 4.97 km/h

Step-by-step explanation:

Given that the time taken by the aircraft = 5 hours

Let the distance covered = X

Speed = V

Time taken by the ship = 7 hours

Let the distance covered = Y

Speed = 8V

Since it is 8 km/h faster than the aircraft carrier.

Using the speed formula

Speed = distance/time

Distance = speed × time

For the aircraft;

X = 5V

For the ship:

Y = 7 × 8V = 56V

Since they are 303km distance apart,

X + Y = 303

5V + 56V = 303

61V = 303

V = 303/61

V = 4.97 km/h

Aircraft carrier speed = 4.97 km/h

5 0

3 years ago

Completely factor each of the given polynomials. If a polynomial cannot be factored, write "not factorable."

Art [367]

4x^2+23x+15=0
Δ=23^2-4*4*15=529-240=289

x1= (-23+17)/8=-6/8= -3/4
x2= (-23-17)/8=-5

4x^2+23x+15= 4(x+3/4)(x+5)=(4x+3)(x+5)

6 0

3 years ago

What is the sum of the first 40 even intergers? A.1720 B.1840 C.1580 D.1640

3241004551 [841]

Answer:

if so 2130.12548...x^2

THATS BOT REALLY THE ANSWER.. I JUST REALLLY NEED TO LVL UP.. IM SO SORRY :(

Step-by-step explanation:

4 0

3 years ago

A rectangular prism is 9.8 inches long, 20 inches wide, and 5 inches high. what is the surface area of the rectangular prism?

Sliva [168]

The prism has 6 faces: 2(length·width), 2(length·heighth), & 2(width·heighth)
2(length · width) = 2(9.8 · 20) = 2(196) = 392 in²
2(length · heighth) = 2(9.8 · 5) = 2(49) = 98 in²
2(width · heighth) = 2(20 · 5) = 2(100) = 200 in²

The sum of the faces is the surface area (S.A.)
392 in² + 98 in² + 200 in² = 690 in²

3 0

3 years ago

Crash testing is a highly expensive procedure to evaluate the ability of an automobile to withstand a serious accident. A simple

polet [3.4K]

Answer:

95% confidence interval for the difference in the proportion is [-0.017 , 0.697].

Step-by-step explanation:

We are given that a simple random sample of 12 small cars were subjected to a head-on collision at 40 miles per hour. Of them 8 were "totaled," meaning that the cost of repairs is greater than the value of the car.

Another sample of 15 large cars were subjected to the same test, and 5 of them were totaled.

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

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of small cars that were totaled = $\frac{8}{12}$ = 0.67

$\hat p_2$ = sample proportion of large cars that were totaled = $\frac{5}{15}$ = 0.33

$n_1$ = sample of small cars = 12

$n_2$ = sample of large cars = 15

$p_1$ = population proportion of small cars that are totaled

$p_2$ = population proportion of large cars that were totaled

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

So, 95% confidence interval for the difference between population population, ( $p_1-p_2$ ) 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 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ < $p_1-p_2$ < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ) = 0.95

95% confidence interval for $p_1-p_2$ = [ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ]

= [ $(0.67-0.33)-1.96 \times {\sqrt{\frac{0.67(1-0.67)}{12}+\frac{0.33(1-0.33)}{15} } }$ , $(0.67-0.33)+1.96 \times {\sqrt{\frac{0.67(1-0.67)}{12}+\frac{0.33(1-0.33)}{15} } }$ ]

= [-0.017 , 0.697]

Therefore, 95% confidence interval for the difference between proportions l and 2 is [-0.017 , 0.697].

6 0

3 years ago