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vampirchik [111]
2 years ago
14

HOW AM I SUPPOSED TO DO THIS IF YOU DONT KNOW HOW TO DO THIS DONT HELP AN GET ME SOMEONE THAT WILL someone help me

Mathematics
1 answer:
Romashka-Z-Leto [24]2 years ago
6 0
Just take each number in the first table and divide it by the total then x by 100

Eg the first one 24/120 x 100 = 20%
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Help Q-Q 7/10c =4 1/5
shepuryov [24]

\huge\mathfrak\green{answer}

= q -   \frac{q7}{10} c =  \frac{41}{5}

\red{multiply}

= q -  \frac{cq7}{10} =  \frac{41}{5}

\red{subtract \: q \: from \: both \: sides}

= q -  \frac{cq7}{10}  - q =  \frac{41}{5} - q

\red{now \: simplify}

=  -  \frac{cq7}{10}  =  \frac{41}{5}  - q

\red{multiply \: 10 \: both \: sides}

= 10( -  \frac{cq7}{10} ) = 10. \frac{41}{5}  - 10q

\red{simplify}

=  - cq7 = 82 - 10q

\red{now \: divide}

=  \frac{ - cq7}{ - q7}  -  \frac{82}{ - q7}  -  \frac{10q}{ - q7}

\red{simplify}

= c =   - \frac{82 - 10q}{q7} (q≠0)

Brainliest? :) (I'd really appreciate if you mark me as brainliest)

\huge\mathfrak\green{thank \: you}

5 0
3 years ago
Read 2 more answers
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
Read the sentence below and answer the following question: Roy was very careful to cleave the apple into even portions. What doe
irga5000 [103]
To separate into parts
5 0
3 years ago
-4(x+3)^2+10 convert it to standard from
Alenkasestr [34]
-4x+154 should be the answer
6 0
3 years ago
a car can drive 600 miles on 18 gallons of gasoline. How many miles can a car drive on 15 gallons of gasoline​
PolarNik [594]

Answer:

Use the cross technique if it’s name like that in English but , I wish you can understand it from what I‘ve written .

3 0
2 years ago
Read 2 more answers
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