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GarryVolchara [31]

3 years ago

6

The temperature in an oven changes from 350 degrees to 362% What is the percent increase in temperature, to the nearest tenth of

a percent?

1 answer:

Darya [45]3 years ago

3 0

Answer:

3.43%

Step-by-step explanation:

362-350=12

12:350*100 =

(12*100):350 =

1200:350 = 3.43

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Suppose a baseball team has 14 players on the roster who are not members of the pitching staff. Of those 14 players, assume that

Nadusha1986 [10]

Answer: Our required probability would be 0.70.

Step-by-step explanation:

Since we have given that

Number of players = 14

Number of players have recently taken a performance enhancing drug = 3

Number of players have not recently taken a performance enhancing drug = 14-3=11

Number of members chosen randomly = 5

We need to find the probability that at least one of the tested players is found to have taken a performance enhancing drug.

P(Atleast 1) = 1 - P(none is found to have taken a performance enhancing drug)

So, P(X≥1)=1-P(X=0)

$P(X\geq 1)=1-^5C_0(\dfrac{11}{14})^5\\\\P(X\geq 1)=1-(0.786)^5\\\\P(X\geq 1)=0.70$

Hence, our required probability would be 0.70.

7 0

3 years ago

7b ÷ 12 = 4.2 What is b?

saw5 [17]

$\frac{7b }{12} = 4.2 \ \ /*12 \\ \\7b=4.2*12\\ \\7b= 50.4 \ \ \/:7\\ \\\frac{7}{7} \ b= \frac{50.4 }{7}\\ \\b= 7.2$

5 0

2 years ago

Read 2 more answers

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago

What is the value of x?<br> 76+71=124

Kisachek [45]

Answer:

48

Step-by-step explanation:

x=48

6 0

2 years ago

Read 2 more answers

What is the range of the function y = -x2 + 1?

meriva

<span><span>y = 2 + 2sec(2x) The upper part of the range will be when the secant has the smallest positive value up to infinity. The smallest positive value of the secant is 1 So the minimum of the upper part of the range of y = 2 + 2sec(2x) is 2 + 2(1) = 2 + 2 = 4 So the upper part of the range is [4, ) The lower part of the range will be from negative infinity up to when the secant has the largest negative value. The largest negative value of the secant is -1 So the maximum of the lower part of the range of y = 2 + 2sec(2x) is 2 + 2(-1) = 2 - 2 = 0 So the lower part of the range is (, 0]. Therefore the range is (, 0] U [4, ) </span>
</span>

3 0

3 years ago