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Step2247 [10]
3 years ago
10

The complement of 4 heads in the toss of 4 coins is

Mathematics
1 answer:
Fofino [41]3 years ago
4 0

Answer:

b i believe the answer is

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Help me pleeeeeeeeeaze
kobusy [5.1K]

a(x+3) < 5x+15-x\\\\ax+3a < 5x+15-x\\\\ax+3a < 4x+15\ \ \ |-3a\\\\ax < 4x+15-3a\ \ \ \ |-4x\\\\ax-4x < 15-3a\\\\(a-4)x < 15-3a\\\\\text{The solution is all real numbers if}\ a-4=0\ \ \ |+4\\\\\boxed{a=4}\\\\\text{Check:}\\\\4(x+3) < 5x+15-x\\4x+12 < 4x+15\ \ \ \ |-4x\\12 < 15\ TRUE\to x\in\mathbb{R}

5 0
3 years ago
PLEASE HELP
Crank

Answer:

Listeriosis is an infection caused by the bacteria Listeria monocytogenes. People become infected by eating foods contaminated with the bacteria. Listeria may infect many different sites in the body, such as the brain, spinal cord membranes, or the bloodstream

Step-by-step explanation:

People with invasive listeriosis usually report symptoms starting 1 to 4 weeks after eating food contaminated with Listeria

5 0
3 years ago
Read 2 more answers
Find the arc length of the partial circle. <br>length is 1.<br>**See attached photo**<br>​
KatRina [158]

Arc length of the quarter circle is 1.57 units.

Solution:

Radius of the quarter circle = 1

Center angle (θ) = 90°

To find the arc length of the quarter circle:

$\text{Arc length}=2 \pi r\left(\frac{\theta}{360^\circ}\right)

                 $=2 \times 3.14  \times 1\left(\frac{90^\circ}{360^\circ}\right)

                $=2 \times 3.14  \times 1\left(\frac{1}{4}\right)

Arc length = 1.57 units

Arc length of the quarter circle is 1.57 units.

5 0
3 years ago
An insurance company has 25,000 automobile policy holders. If the yearly claim of a policy holder is a random variable with mean
telo118 [61]

Answer:

P(T>8300000)=1-P(T

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Data given

n = 25000 represent the automobile policy holders

\mu= 320 represent the population mean

\sigma =540 represent the population standard deviation

Let T the variable that represent the total of interest on this case. We can assume that the random variable for an individual policy holder is given by:

X\sim N(\mu = 540, \sigma=540)

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Solution to the problem

First we need to find the distribution for the random variable T like this:

\bar X = \frac{\sum_{i=1}^n x_i}{n}

And the total T is given by:

T=\sum_{i=1}^n X_i =n \bar X

We can find the expected value, variance and deviation for this random variable like this:

E(T)= n E(\bar X) = n \mu = 25000*320=8000000

Var(T)= Var(n\bar X)= n^2 Var(\bar X) = n^2 \frac{\sigma^2}{n}=n \sigma^2 =25000*(540^2)=7290000000

Sd(T)=\sqrt{7290000000}=85381.497

And we are interested on this probability:

P(T>8300000)

And we can use the Z score formula given by:

Z=\frac{T-E(T)}{\sigma_T}

P(T>8300000)=1-P(T

6 0
3 years ago
How many solutions does this equation have
Stells [14]

Answer:

I m not sure, I can't see

Step-by-step explanation:

8 0
3 years ago
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