Literally have no clue how to do the heart beat please help
Literally have no clue how to do the heart beat - 1

Answers

Answer 1
Answer:

Step-by-step explanation:

To find the rate, divide the number of heartbeats by the number of seconds.

You: 22 beats / 20 seconds = 1.1 beats per second.

Friend: 18 beats / 15 seconds = 1.2 beats per second.


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Lin rode her bike 2 miles in 8 minutes. She rode at a constant speed. Complete the table to show the distance traveled in 1 minute at this speed

Answers

quarter mile, means she gets 1 mile every 4 minutes, there’s 4 quarters when it comes to every 1 minute, so every minute is a quarter mile

Answer:

4 miles in 1 min

Step-by-step explanation:

Please let me know. This is 2nd grade math.'how is exercise 5 different from exercise 6 explain.'

Answers

Answer:

Show a photo

Step-by-step explanation:

we can't see what you're talking about

Can you show a photo of your question? Because I don't know how to answer that.

What statement best describes a trapezoid? *1 point
a. All sides are equal and opposite sides are parallel
b. Exactly one pair of parallel lines
c. Has 2 sets of parallel lines and four 90 degree angles
d. Has 2 equal sides and 3 acute angles

Answers

Answer:b

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

because it has two parallel side and and other are just lines but being quadrilateral its sum is 360

The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives 16 years; thestandard deviation is 1.7 years.
Use the empirical rule (68 - 95 - 99.7%) to estimate the probability of a gorilla living longer than 14.3
years.


Percent % pls

Answers

the probability of a gorilla living longer than 14.3 years is 83.9%

Given :

The lifespans of gorillas in a particular zoo are normally distributed

Mean is 16 years  and standard deviation is 1.7 years

Empirical rule diagram is attached below

We need to find the probability of a gorilla living longer than 14.3

Lets find out 14.3 lies in which standard deviation on left  or right

mean is 16

mean - standard \; deviation =16-1.7=14.3

14.3 lies on first standard deviation on left of mean 16

So we find out the area that covers after 14.3

The area after 14.3 is 34+34++13.5+2.4=83.9

the probability of a gorilla living longer than 14.3 years is 83.9%

Learn more : brainly.com/question/14280851

Final answer:

The probability of a gorilla living longer than 14.3 years is estimated to be 81.2% using the empirical rule.

Explanation:

To estimate the probability of a gorilla living longer than 14.3 years, we can use the empirical rule, also known as the 68-95-99.7% rule. According to this rule, for a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.

The average lifespan of gorillas in this zoo is 16 years, with a standard deviation of 1.7 years. To estimate the probability of a gorilla living longer than 14.3 years, we need to calculate the z-score. The z-score formula is:

z = (x - μ) / σ

where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.

Plugging in the values, we have:

z = (14.3 - 16) / 1.7

Solving this, we get a z-score of -0.88. Using a z-table or a calculator, we can find that the probability of a gorilla living longer than 14.3 years is approximately 0.812, or 81.2%.

Learn more about Estimating probability here:

brainly.com/question/28498058

#SPJ12

A study is being conducted to compare the average training time for two groups of airport security personnel: those who work for the federal government and those employed by private security companies. From a random sample of 12 government-employed security personnel, average training time was 72 hours, with a sample standard deviation of 8 hours. In a random sample of 16 privately employed security personnel, training time was 65.4 hours, with a sample standard deviation of 12.3 hours. Assume that training time for each group is normally distributed. Use the following notations:μ1: The mean training time for the population of airport security personnel
employed by the federal government.
μ2: The mean training time for the population of airport security personnel
employed by private security companies.
The goal of the statistical analysis is to determine whether the sample data support the hypothesis that average training time for government-employed security personnel is higher than those employed by private security companies.
1. What is the null hypothesis H0?
Select one:
a. μ1- μ2 <= 0
b. μ1- μ2 < 0
c. μ1- μ2 =/ 0
d. μ1- μ2 > 0
2. What is the alternative hypothesis Ha?
Select one:
a. μ1- μ2 > 0
b. μ1- μ2 <= 0
c. μ1- μ2 = 0
d. μ1- μ2 >= 0

Answers

Answer:

1.a. H₀: μ₁ - μ₂ ≤ 0

2.b. H₁: μ₁ - μ₂ > 0

Step-by-step explanation:

Hello!

The objective is to compare the average training time for two groups of airport security personnel.

Group 1: Security personnel that works for the federal government

n= 12

X[bar]= 72 hs

S= 8hs

Group 2: Security personnel from private companies

n= 16

X[bar]= 65.4 hs

S= 12.3 hs

The goal of the analysis is to test if the average training time for government-employed security personnel is higher than those employed by private security companies, symbolically: μ₁ > μ₂

The null and alternative hypotheses are complementary and exhaustive.

The null hypothesis always represents the "no change situation" and therefore always carries the = symbol. Generally, the researcher's claim is stated in the alternative hypothesis.

With all this in consideration, the hypotheses for this experiment are:

H₀: μ₁ ≤ μ₂

H₁: μ₁ > μ₂

I hope this helps!

Each side of a square is increasing at a rate of 4 cm/s. At what rate (in cm2/s) is the area of the square increasing when the area of the square is 25 cm2

Answers

Answer:

The area of the square is increasing at a rate of 40 square centimeters per second.

Step-by-step explanation:

The area of the square (A), in square centimeters, is represented by the following function:

A = l^(2)(1)

Where l is the side length, in centimeters.

Then, we derive (1) in time to calculate the rate of change of the area of the square ((dA)/(dt)), in square centimeters per second:

(dA)/(dt) = 2\cdot l \cdot (dl)/(dt)

(dA)/(dt) = 2\cdot √(A)\cdot (dl)/(dt)(2)

Where (dl)/(dt) is the rate of change of the side length, in centimeters per second.

If we know that A = 25\,cm^(2) and (dl)/(dt) = 4\,(cm)/(s), then the rate of change of the area of the square is:

(dA)/(dt) = 2\cdot \sqrt{25\,cm^(2)}\cdot \left(4\,(cm)/(s) \right)

(dA)/(dt) = 40\,(cm^(2))/(s)

The area of the square is increasing at a rate of 40 square centimeters per second.