20 Pts + Brainliest for best/first answer

Someone please help (show work)

Mnenie [13.5K]

Answer:

See answers below :)

Step-by-step explanation:

-2(2 - 5) + 10 + (-5) =

-2(-3) + 10 - 5 =

6 + 10 - 5 =

11

8 - 2(9 - 5)^2 - 1 =

8 - 2(4)^2 - 1 =

8 - 2(16) - 1 =

8 - 32 - 1 =

-25

8 0

3 years ago

5y + 3 = 8y − 5 + 2y

DIA [1.3K]

Answer:

$\fbox{y = 8/5 }$

Explanation:

[to solve for y, first use the properties of equality to simplify the equation]

5y + 3 = 8y − 5 + 2y

5y + 3 = (8y + 2y) – 5

[regroup the like terms of y together: commutative property of equality ; adding in a different order will still give you the same result]

5y + 3 = 10y – 5

[combine like terms]

5y + 3 = 10y – 5

$\: \: \: \: \: \: -3 \: \: \: \: \: \: \: \: \: \: \: -3$

[subtract 3 from both sides in order to eliminate the constant term on the left side: subtraction property of equality]

5y = 10y – 8

–10y –10y

[subtract 10 from both sides in order to eliminate the variable term: subtraction property of equality]

-5y = -8

÷(-5) ÷(-5)

[divide both sides by -5 to cancel out the coefficient of y: division property of equality]

y = 8/5

7 0

2 years ago

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Coach Johnson need to buy shin guards for junior varsity and varsity basketball players the following package of shin guards are

Firdavs [7]

Answer:

$1.50

Step-by-step explanation:

To find the best deal, we first want to consider price per unit. To find the price per unit, we divide the price by the number of units. For the 10 shin guard package, our price is 14.50, and we divide that by 10 to get 1.45 . Therefore, our unit price for the package of 10 is $1.45 per shin guard. Similarly, we can find the unit price for the package of 15 to be 22.5/15 = $1.50 per shin guard. As 1.50 is greater than 1.45, the lowest unit price is for the package of 10.

The question is asking for us to compare the prices if we bought 30 of each. For the package of 10, we get 1.45*30 (as we're buying 30 for 1.45) = 43.5 as the total price, and for the package of 15, we get 1.5*30 = 45 as our price. As 15 and 10 are both factors of 30, we don't need to worry about converting it back into packages of 10/15. The difference between buying 30 at the lowest and highest unit price is therefore (45-43.5)=1.5 dollars

4 0

3 years ago

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A survey report states that 70% of adult women visit their doctors for a physical examination at least once in two years. If 20

irakobra [83]

Answer:

a) 0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

b) 0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

Step-by-step explanation:

For each women, there are only two possible outcomes. Either they visit their doctors for a physical examination at least once in two years, or they do not. The probability of a woman visiting their doctor at least once in this period is independent of any other women. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

70% of adult women visit their doctors for a physical examination at least once in two years.

This means that $p = 0.7$

20 adult women

This means that $n = 20$

(a) Fewer than 14 of them have had a physical examination in the past two years.

This is:

$P(X < 14) = 1 - P(X \geq 14)$

In which

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 14) = C_{20,14}.(0.7)^{14}.(0.3)^{6} = 0.1916$

$P(X = 15) = C_{20,15}.(0.7)^{15}.(0.3)^{5} = 0.1789$

$P(X = 16) = C_{20,16}.(0.7)^{16}.(0.3)^{4} = 0.1304$

$P(X = 17) = C_{20,14}.(0.7)^{17}.(0.3)^{3} = 0.0716$

$P(X = 18) = C_{20,18}.(0.7)^{18}.(0.3)^{2} = 0.0278$

$P(X = 19) = C_{20,19}.(0.7)^{19}.(0.3)^{1} = 0.0068$

$P(X = 20) = C_{20,20}.(0.7)^{20}.(0.3)^{0} = 0.0008$

So

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1916 + 0.1789 + 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.6079$

$P(X < 14) = 1 - P(X \geq 14) = 1 - 0.6079 = 0.3921$

0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

(b) At least 17 of them have had a physical examination in the past two years

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

From the values found in item (a).

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.107$

0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

6 0

2 years ago

Can you please help me with the sheet??

OLga [1]

1. 25 minutes
2. 41 minutes
3. 21 wheels
4. 7
5. 160
6. 415 miles

7 0

3 years ago

Read 2 more answers