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2 years ago

Thx to all that helped me!!! here is my last question

Mathematics

2 answers:

Virty [35]2 years ago

7 0

Answer:

her younger sister's age: (m - 3)years , her uncles age ( 5m )

sum of their ages : m + m - 3 + 5m = 7m - 3

Leviafan [203]2 years ago

4 0

Answer:

cousin: m-3

uncle: 5m

together: 6m-3

Step-by-step explanation:

if her cousin is 3 years younger, they would be m-3. since their uncle is her age times 5, he would be 5m. add those and you get 5m+m-3. combine the m's (and get 6m), then add the -3 to get 6m-3

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Hey can anyone help me with this question please and thank you !

Inessa [10]

Answer:

(6x-12)+(8x-4)=180

14x-16=180

14x=196

x=14

Step-by-step explanation:

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2 years ago

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Ede4ka [16]

Answer:

Commutative property of multiplication

Step-by-step explanation:

Commutative property of multiplication basically means that the order in which you multiply numbers does not matter.

For exampe:

3 x 4 = 4 x 3

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In your case:

5 x 7 x 4 = 5 x 4 x 7

140 = 140

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3 years ago

48 cookies 25% increase

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Answer:

Step-by-step explanation:

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2 years ago

1000ml:300ml to its simplest form

Diano4ka-milaya [45]

<h2>HI MATE YOUR ANSWER SHOULD BE 10/3</h2>

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

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3 years ago