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2 years ago

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Ms. Estrada plans to take 36 chocolate chip cookies to sell at a bake sale. She wants to bake enough to keep at least 24 cookies

for her cookie jar at home.
Let x represent how many chocolate chip cookies Ms. Estrada will bake. Which inequality describes the problem?

1 answer:

Kruka [31]2 years ago

4 0

Answer:

36-x≥24

Step-by-step explanation:

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A sample of 1800 computer chips revealed that 53% of the chips do not fail in the first 1000 hours of their use. The company's p

charle [14.2K]

Answer:

So the value of the test statistic is -0.85.

Step-by-step explanation:

We know that a sample of 1800 computer chips revealed that 53% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 54% of the chips do not fail in the first 1000 hours of their use.

We get that:

$n=1800\\\\p_1=53\%=0.53\\\\p_2=54\%=0.54\\$

We calculate the standar deviation:

$\sigma=\sqrt{\frac{0.53 \cdot (1-0.53)}{n}}=\sqrt{\frac{0.53 \cdot 0.47}{1800}}=0.01176$

We calculate the value of the test statistic:

$z=\frac{p_1-p_2}{\sigma}=\frac{0.53-0.54}{0.01176}=-0.85$

So the value of the test statistic is -0.85.

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3 years ago

What is the answer to 1.75g x 1000mg/1g?

Lerok [7]

1750mg in 1.75g just move the decimal place over three places. so its 1750mg

8 0

3 years ago

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.

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3 years ago

Help me with this question please

docker41 [41]

7x/x-4 * x/x+7

mutiply the numerators together

(7x)(x)= 7x^ 2

mutiply the denominators together

(x-4)(x+7)

(x)(x)(7)(x)= x^2+7x

(-4)(x)(-4)(7)= -4x-28

x^2+7x-4x-28

x^2+3x-28

Answer:

7x^2/x^ 2+3x-28

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3 years ago

The mean of 67004, 67004, 67005, 67005, 67005, 67005, 67006, 67006

romanna [79]

Answer:

the answer is 59467

<h2><em>hope it will help you</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em></h2>

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