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2 years ago

7

Add O used 6 cups of whole wheat flour and eggs we flower and ax cups of white flour in the recipe what is the equation that can

be used to find the value of Y the total amount of flour that adult used in the recipe and what are the constraints and the values of X and Y

1 answer:

Naya [18.7K]2 years ago

7 0

Answer:

6x+y

Step-by-step explanation:

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9) -4(r+ 2) = -16<br> B) (2)<br> C) (-10)<br> A) (10)<br> What is the answer

GenaCL600 [577]

Answer:

$- 4r - 8 = - 16 \\ - 4r = - 8 \\ r = 2$

The answer option is B.

Good luck.

Intelligent Muslim.

3 0

3 years ago

the telephone company offers two billing plans for local calls. Plan 1 charges $29 per month for unlimited calls and Plan 2 char

kodGreya [7K]

Answer:

Step-by-step explanation:

Plan 2: x= number of months

y= the cost per call

19x + 0.04y

19(4) + 0.04(20)

76 + 0.8

76.8

So the total will be $76.8

4 0

3 years ago

1.) In a the heavy weight division boxing match, Mike throws 270 punches in 9 rounds. What is the unit rate of punches per round

Rom4ik [11]

30 punches each round...
i think :)
just divide the punches by the rounds hope this helps

6 0

3 years ago

Random samples of size 64 are taken from an infinite population whose mean and standard deviation are 24 and 6.4, respectively.

umka2103 [35]

Answer:

Mean 24, standard error 0.8

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation, also called standard error $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 24, \sigma = 6.4, n = 64$

What are the mean and the standard error of the sample mean?

By the Central Limit Theorem, mean 24 and standard error $s = \frac{6.4}{\sqrt{64}} = 0.8$

7 0

2 years ago

If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.

Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have $S_1=3\times2^{1-1}+2(-1)^1=3-2=1$ , which agrees with the initial value <em>S</em>₁ = 1.

• Assume the closed form is correct for all <em>n</em> up to <em>n</em> = <em>k</em>. In particular, we assume

$S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}$

and

$S_k=3\times2^{k-1}+2(-1)^k$

We want to then use this assumption to show the closed form is correct for <em>n</em> = <em>k</em> + 1, or

$S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}$

From the given recurrence, we know

$S_{k+1}=S_k+2S_{k-1}$

so that

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)$

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}$

$S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)$

$S_{k+1}=3\times2^k-2(-1)^k$

$S_{k+1}=3\times2^k+2(-1)(-1)^k$

$\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}$

which is what we needed. QED

6 0

2 years ago