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2 years ago

Cashews cost $6.75/lb. and Jeanna can spend at most $30 for them. How many pounds can she buy?

Mathematics

2 answers:

Vika [28.1K]2 years ago

6 0

Answer:

4.44 lb

Step-by-step explanation:

$6.75 ---> 1 lb

$1 ---> (1/6.75) lb

$30 ---> [$30* (1/6.75)] lb

$30 ---> 4.44

So she can buy most 4.44 lb of pounds

Step-by-step explanation:

pls branliest to me he/she have 210 me onley 4

Harman [31]2 years ago

4 0

Answer:

4.44 lb

Step-by-step explanation:

$6.75 ---> 1 lb

$1 ---> (1/6.75) lb

$30 ---> [$30* (1/6.75)] lb

$30 ---> 4.44

So she can buy most 4.44 lb of pounds

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sin(A) = $\frac{6}{10} = \frac{3}{5}$

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Could anyone tell me if I’m right. This is a gradient of a line question

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Answer:

See Below (it is correct)

Step-by-step explanation:

First point given ---- A(-3,0)

2nd point given ----- B(4,0)

The slope is the "change in quantity y" divided by "change in quantity x"

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So, slope would be:

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3 years ago

Consider the quadratic equation x? = 4x - 5. How many solutions does the equation have? A The equation has one real solution. Th

anzhelika [568]

Answer:

{-1, 5} (two real solutions).

Step-by-step explanation:

On the left side we want x^2, not x?

We need to rewrite x? = 4x - 5 in standard form, that is, as a quadratic with x terms in decreasing powers of x:

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4 0

3 years ago

10. If y = -2x + 3, find y when x = -3.<br> a) 12<br> b) -2

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Answer:

Step-by-step explanation:

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5 0

3 years ago

Ocean fishing for billfish is very popular in the Cozumel region of Mexico. In the Cozumel region about 41% of strikes (while tr

Klio2033 [76]

Answer:

a) 59.10% probability that 12 or fewer fish were caught.

b) 99.74% probability that 5 or more fish were caught.

c) 58.84% probability that between 5 and 12 fish were caught.

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 29, p = 0.41$

$\mu = E(X) = np = 29*0.41 = 11.89$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = 2.6486$

Find the following probabilities.

a) 12 or fewer fish were caught.

Using continuity correction, this is $P(X \leq 12 + 0.5) = P(X \leq 12.5)$ , which is the pvalue of Z when X = 12.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{12.5 - 11.89}{2.6486}$

$Z = 0.23$

$Z = 0.23$ has a pvalue of 0.5910

59.10% probability that 12 or fewer fish were caught.

b) 5 or more fish were caught.

Using continuity correction, this is $P(X \geq 5 - 0.5) = P(X \geq 4.5)$ , which is 1 subtracted by the pvalue of Z when X = 4.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4.5 - 11.89}{2.6486}$

$Z = -2.79$

$Z = -2.79$ has a pvalue of 0.0026

1 - 0.0026 = 0.9974

99.74% probability that 5 or more fish were caught.

c) between 5 and 12 fish were caught.

Using continuity correction, this is $P(5 - 0.5 \leq X \leq 12 + 0.5) = P(4.5 \leq X \leq 12.5)$ , which is the pvalue of Z when X = 12.5 subtracted by the pvalue of Z when X = 4.5. So.

From a), when X = 12.5, Z has a pvalue of 0.5910

From b), when X = 4.5, Z has a pvalue of 0.0026.

0.5910 - 0.0026 = 0.5884

58.84% probability that between 5 and 12 fish were caught.

8 0

4 years ago