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3 years ago

Divya also wants to make fruit smoothies for dessert. She purchases

Mathematics

2 answers:

MAVERICK [17]3 years ago

6 0

Answer:

Divya purchases a container of rice milk at $3.49

two pounds of strawberries, $4.99 each; $9.98

six bananas at $0.75 each; $4.50

Therefore,

$3.49+$9.98+$4.50=$17.97

Jane spends a total of $17.97 on dessert.

Step-by-step explanation:

A cheerful teen willing to help,

stay salty,

be positive,

hope you find this useful...

Slav-nsk [51]3 years ago

5 0

Answer:

$17.97

Step-by-step explanation:

add everything up

- bananas: 6(0.75) = $4.50

- rice milk: $3.49

- strawberries: 2(4.99) = $9.98

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To decrease the impact on the environment, factory chimneys must be high enough to allow pollutants to dissipate over a larger a

Komok [63]

Answer:

The probability hat the sample mean height for the 40 chimneys is greater than 102 meters is 0.1469.

Step-by-step explanation:

Let the random variable X be defined as the height of chimneys in factories.

The mean height is, μ = 100 meters.

The standard deviation of heights is, σ = 12 meters.

It is provided that a random sample of n = 40 chimney heights is obtained.

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Then, the mean of the distribution of sample means is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the distribution of sample means is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

Since the sample selected is quite large, i.e. n = 40 > 30, the central limit theorem can be used to approximate the sampling distribution of sample mean heights of chimneys.

$\bar X\sim N(\mu_{\bar x},\ \sigma^{2}_{\bar x})$

Compute the probability hat the sample mean height for the 40 chimneys is greater than 102 meters as follows:

$P(\bar X>102)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}})>\frac{102-100}{12/\sqrt{40}})$

$=P(Z>1.05)\\=1-P(Z$

*Use a z-table fr the probability.

Thus, the probability hat the sample mean height for the 40 chimneys is greater than 102 meters is 0.1469.

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I think you meant to say

$\displaystyle \lim_{t\to2}\frac{t^4-6}{2t^2-3t+7}$

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Since both the numerator and denominator are continuous at t = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)}$

Because these expressions are continuous at t = 2, we can compute the limits by evaluating the limands directly at 2:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)} = \frac{2^4-6}{2\cdot2^2-3\cdot2+7} = \boxed{\frac{10}9}$

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