I need help w number 6

Mathematics

2 answers:

Y_Kistochka [10]3 years ago

6 0

Answer:

I would say that it is A

Step-by-step explanation:

Oduvanchick [21]3 years ago

5 0

Answer:

Step-by-step explanation:

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What is an equation in pointelope form for the line that passes through the points (4. – 1) and (-3,0)

bulgar [2K]

Answer:

1/3

Step-by-step explanation:

2nd one

5 0

2 years ago

Joe went on holiday to Spain.

Hunter-Best [27]

Answers:

a. £2,090
b. My answer would be less because my total in euros would be divided by a greater number when converting euros to pounds.

Work for part a:

Apartment: 560•3= 1680 euros

Car: 20.16•15= 302.4 euros

1680+302.4=1982.4

1982.4/1.12 = £x/1

1982.4 / 1.12= £1,770

£1,770 + £320 = £2,090 total

Work for part b:
Example:
(instead of 1.12 I put 2 to see what would happen with a greater number)
1982.4 / 2 = 991.2

991.2 is less than 1,770 (the answer I got with 1.12)

Let me know if you need any clarification! :)

7 0

3 years ago

Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave

Colt1911 [192]

Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

$P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057}$

Part B:

</span>The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

</span> $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$