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amid [387]
3 years ago
14

Pls help will be nice

Mathematics
2 answers:
Oliga [24]3 years ago
6 0
First one have a good day
ʕ•ᴥ•ʔʕ•ᴥ•ʔ
Naily [24]3 years ago
4 0

Answer:

first one(|-1-4|) and second one (|4-(-1)|)

Step-by-step explanation:

distance = 5 and both are equals to 5 cause of absolute value. Third and four ones result is 3.

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Maryanne sells cruise vacations.
grandymaker [24]

Answer:

fcriohewdsajiodsx

Step-by-step explanation:

oing this for something! tewsting

4 0
3 years ago
Read 2 more answers
Which of the following are true?
matrenka [14]
The correct answer would be a.
7 0
3 years ago
In circle B with the measure of minor arc AC= 126°, find mŁADC.
Pavel [41]

Answer:

54 degrees

Step-by-step explanation:

First we have to find angle ADC.

DC = 360 - 180 - 126

= 360 - 306

= 54 degrees

Hope it helps love.

;)

7 0
2 years ago
4) A salesperson is paid $60 per week plus $4.50 per sale. This week, the salesperson wants to earn at least $250. How many sale
sasho [114]

。☆✼★ ━━━━━━━━━━━━━━  ☾  

Subtract the $60 out of the total first

250 - 60 = 190

Divide this value by pay per sale:

190 / 4.50 = 42.2222

Thus, they must make 43 sales in order to meet the goal

Have A Nice Day ❤    

Stay Brainly! ヅ    

- Ally ✧    

。☆✼★ ━━━━━━━━━━━━━━  ☾

3 0
3 years ago
PROBLEM SOLVING Scientists conducted aerial surveys of a seal sanctuary and recorded the number x of seals they observed during
Akimi4 [234]

Answer:

Probability that at most 50 seals were observed during a randomly chosen survey is 0.0516.

Step-by-step explanation:

We are given that Scientists conducted aerial surveys of a seal sanctuary and recorded the number x of seals they observed during each  survey.

The numbers of seals observed were normally distributed with a mean of 73 seals and a standard deviation of 14.1 seals.

Let X = <u><em>numbers of seals observed</em></u>

The z score probability distribution for normal distribution is given by;

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

where, \mu = population mean numbers of seals = 73

            \sigma = standard deviation = 14.1

Now, the probability that at most 50 seals were observed during a randomly chosen survey is given by = P(X \leq 50 seals)

            P(X \leq 50) = P( \frac{X-\mu}{\sigma} \leq \frac{50-73}{14.1} ) = P(Z \leq -1.63) = 1 - P(Z < 1.63)    

                                                          = 1 - 0.94845 = <u>0.0516</u>

The above probability is calculated by looking at the value of x = 1.63 in the z table which has an area of 0.94845.

4 0
4 years ago
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