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3 years ago

I need help

Mathematics

2 answers:

kotykmax [81]3 years ago

5 0

Answer:okay send me a picture of your lesson

Step-by-step explanation:

Ksivusya [100]3 years ago

5 0

Answer:

Step-by-step explanation:

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What is 8.1 divided by ( -0.9 ) ?

Ksju [112]

-9 hopefully this helps

8 0

3 years ago

Read 2 more answers

Jackson has a table with a square top and he wants to buy a circular piece of lace that will cover the entire top of the table.

DedPeter [7]

Answer:

226.2

Step-by-step explanation:

I took the test. I hope this helps! :)

6 0

3 years ago

Evaluate the six trigonometric functions of the angle 0

Alla [95]

# Sin θ = 9/15.

# Cos θ = 12/15.

# Cosec θ = 15/9.

# Sec θ = 5/4

Step-by-step explanation:

$by \: using \: pythagorian \: triplets$

${a}^{2} + {b}^{2} = {c}^{2}$

${9}^{2} + {12}^{2} = {c}^{2}$

$81 + 144 = {c}^{2}$

$225 = {c}^{2}$

$c = 15.$

$\sin θ = \frac{9}{15}$

$\cos θ = \frac{12}{15} = \frac{4}{5}$

$\csc θ = \frac{15}{9}$

$\secθ = \frac{5}{4}$

7 0

3 years ago

The mean volume for a bottle of cologne is 4 ounces and the standard deviation is 0.22 ounces. A random sample of 121 bottles is

adelina 88 [10]

Answer:

The standard error of the mean is 0.02

Step-by-step explanation:

Here, we want to calculate the standard error of the mean.

Mathematically;

standard error of mean = SD/ √n

where SD is the standard deviation and n is the number of samples

From the question, SD = 0.22 and n = 121

Inserting this into the equation

Standard error of the mean = 0.22/√121

= 0.22/11 = 0.02

6 0

3 years ago

A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses

Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = average age of the random sample of horses with colic = 12 yrs

$\mu$ = average age of all horses seen at the veterinary clinic = 10 yrs

$\sigma$ = standard deviation of all horses coming to the veterinary clinic = 8 yrs

n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P( $\bar X$ $\geq$ 12)

P( $\bar X$ $\geq$ 12) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{12-10}{\frac{8}{\sqrt{60} } }$ ) = P(Z $\geq$ 1.94) = 1 - P(Z < 1.94)

= 1 - 0.97381 = 0.0262

Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

4 0

3 years ago