Every day, you take the elevator from the basement parking lot that is at level-3. The elevator takes twoseconds per level and you find that it takes 46 seconds to reach your floor.
Which floor do you live on?

Answers

Answer 1

Answer: 23 is the floor which you live on

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Jocelyn and Lorlesha are comparing the size of their villages in the Clash of Clans app. The area of Jocelyn's village is represented by the polynomial, 2w2+10w+12. the area of lorleshas village is represented by the polynomial, 3w2+4w-5, where e represents the width, in meters of their Town hall. part A: find the expression that represents the additional area Jocelyns village.part B: find the expression that represents the combined total area of their villages.
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Plz help i dont understand and if u can help i will mark brainliest!!

Answers

Answer:

Let fertilizer be F

and peat moss be P

since F is proportional to P

F = kP

where, k = constant of proportionality

k = F/P

putting in the values,

k = (3/4) /12

k = 3/4 * 1/12

k = 1/16

Therefore the constant of proportionality is 1/16.

Hey can you please help me posted picture of question

Answers

For this case we have the following equation:
y = 4 (x-2) ^ 2-1
Rewriting we have:
y = 4 (x ^ 2-4x + 4) ^ 2-1
Multiplying the common factor 2 we have:
y = 4x ^ 2-16x + 16-1
Adding the constant term we have:
y = 4x ^ 2-16x + 15
Answer:
y = 4x ^ 2-16x + 15
option D

the answer is D y=4x^2-16x+15

You need to lay communications cable around the perimeter of three offices, each 8' by 10'. How much cable do you need?

Answers

Answer:

108 units

Step-by-step explanation:

We are given that

Length of each office,l=10 units

Breadth of each office=b=8 units

Total number of office=3

We know that

Perimeter of rectangle= $2(l+b)$

Using the formula

Length of cable for 1 office= $2(10+8)=36units$

Length of cable required for 3 offices= $36* 3=108$ units

Hence, the required length of cable=108 units

*An isosceles trapezoid ABCD has bases AD = 17cm, BC = 5cm, and leg AB = 10 cm. A line is drawn through vertex B so that it bisects diagonal AC
and intersect
AD
at point M. Find the area of ΔBDM. What is the area of ABCD?

Answers

The diagram is attached below

The midpoint of AC is E. The ΔMAE is congruent to ΔBCE, and the sides MA = BC = 5 cm. Also MD = 17 -5 = 12 cm

From the figure we can see that the height of the trapezoid is 8cm

The height of trapezoid is the difference between the x axis and the BC that is 8cm

The triangle BDM has base 12 cm and height 8 cm.

Area of triangle = $(1)/(2) * base * height$

A = $(1)/(2) * 12 * 8$ = 48 cm²

2. The area of trapezoid ABCD formula is

A = $(base_1 + base_2)/(2)* height$

base 1= 17cm , base2= 5cm , height = 8cm

A = $(17 + 5)/(2)* 8$

A = 88 cm²

Asked and answered elsewhere.
brainly.com/question/9635743

. In a study of air-bag effectiveness it was found that in 821 crashes of midsize cars equipped with air bags, 46 of the crashes resulted in hospitalization of the drivers (based on data from the highway Loss Data Institute). Using a 0.01 significance level, you need to test the claim that the air-bag hospitalization rate is lower than the 7.8% rate for crashes of mid-size cars equipped with automatic safety belts. What conclusion should you make

Answers

Answer:

P(X ≤ 46 | X~B(821, 0.078)) = 0.00885745584

0.00885... < 0.01

The test statistic of 46 is significant

There is sufficient evidence to reject H₀ and accept H₁

Air bags are more effective as protection than safety belts

Step-by-step explanation:

821 crashes

46 hospitalisations where car has air bags

7.8% or 0.078 probability of hospitalisations in cars with automatic safety belts

α = 0.01 or 1% ← level of significance

One-tailed test

We are testing whether hospitalisations in cars with air bags are less likely than in a car with automatic safety belts;

The likelihood of hospitalisation in a car with automatic safety belts, we are told, is 7.8% or 0.078;

So we are testing if hospitalisations in cars with air bags is less than 0.078;

So, firstly:

Let X be the continuous random variable, the number of hospitalisations from a car crash with equipped air bags

X~B(821, 0.078)

Null hypothesis (H₀): p = 0.078

Alternative hypothesis (H₁): p < 0.078

According to the information, we reject H₀ if:

P(X ≤ 46 | X~B(821, 0.078)) < 0.01

To find P(X ≤ 46) or equally P(X < 47), it could be quite long-winded to do manually for this particular scenario;

If you are interested, the manual process involves using the formula for every value of x up to and including 46, i.e. x = 0, x = 1, x = 2, etc. until x = 46, the formula is:

$P(X = r) = nCr * p^(r) * (1 - p)^(n - r)$

You can find binomial distribution calculators online, where you input n (i.e. the number of trials or 821 in this case), probability (i.e. 0.078) and the test statistic (i.e. 46), it does it all for you, which gives:

P(X ≤ 46 | X~B(821, 0.078)) = 0.00885745584

Now, we need to consider if the condition for rejecting H₀ is met and recognise that:

0.00885... < 0.01

There is sufficient evidence to reject H₀ and accept H₁.

To explain what this means:

The test statistic of 46 is significant according to the 1% significance level, meaning the likelihood that only 46 hospitalisations are seen in car crashes with air bags in the car as compared to the expected number in car crashes with automatic safety belts is very unlikely, less than 1%, to be simply down to chance;

In other words, there is 99%+ probability that the lower number of hospitalisations in car crashes with air bags is due to some reason, such as air bags being more effective as a protective implement than the safety belts in car crashes.

Need help pleaseeee!