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2 years ago

6

On the previous screen, you said the daily recommended amount of vitamin C is `80` milligrams. Yosef drank `4` ounces of orange

juice. What percent of the daily recommended amount of vitamin C is this?A `16`-ounce bottle of orange juice contains `200` milligrams of vitamin C, which is `250\%` of the daily recommended amount (for adults).

1 answer:

Sindrei [870]2 years ago

5 0

Answer: 62.5%

Step-by-step explanation:

We can just divide 250% by 4 since Yosef drank 1/4th of the bottle.

250 / 4 = 62.5

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Feet

Step-by-step explanation:

Inches and centimeters are really small, so it wouldn't be appropriate to measure it in those units. Miles are really long, so that to would not be appropriate.

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3 years ago

The average yield of the standard variety off soybeans per acre on a farm in a region is 48.8 bushels per acre, with a standard

I am Lyosha [343]

Answer:

$z=\frac{53.4-48.8}{\frac{12}{\sqrt{36}}}=2.3$

$p_v =P(Z>2.3)=1-P(Z$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 48.8 at 1% of signficance.

Step-by-step explanation:

1) Data given and notation

$\bar X=53.4$ represent the sample mean

$\sigma=12$ represent the population standard deviation assumed

$n=36$ sample size

$\mu_o =48.8$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is higher than 48.8, the system of hypothesis would be:

Null hypothesis: $\mu \leq 48$

Alternative hypothesis: $\mu > 48$

Since we assume that know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{53.4-48.8}{\frac{12}{\sqrt{36}}}=2.3$

4)P-value

Since is a right tailed test the p value would be:

$p_v =P(Z>2.3)=1-P(Z$

5) Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 48.8 at 1% of signficance.

3 0

3 years ago

Hi- I need help please to get my HS diploma...did not graduate :(

andrew-mc [135]

The remainder from the division of the algebraic equation is -53/8.

<h3>What is the remainder of the algebraic expression?</h3>

The remainder of the algebraic expression can be determined by using the long division method.

Given that:
$\mathbf{f(x) = \dfrac{x^3 - 6x^2 + 3x - 1}{2x-3}}$

where:

The divisor = 2x -3

Using the long division method, we have:

$\mathbf{= \dfrac{x^2}{2} +\dfrac{-\dfrac{9x^2}{2}+3x -1 }{2x-3}}$

$\mathbf{= \dfrac{x^2}{2}-\dfrac{9x}{4} +\dfrac{-\dfrac{-15x}{4}-1 }{2x-3}}$

$\mathbf{= \dfrac{x^2}{2}-\dfrac{9x}{4} -\dfrac{15}{8}+\dfrac{-\dfrac{53}{8} }{2x-3}}$

$\mathbf{= \dfrac{x^2}{2}-\dfrac{9x}{4} -\dfrac{15}{8}-\dfrac{53 }{8(2x-3)}}$

Therefore, we can conclude that the remainder is -53/8.

Learn more about the division of algebraic equations here:

brainly.com/question/4541471

#SPJ1

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2 years ago

Help please i don’t know how to do it

Norma-Jean [14]

Answer:

Step-by-step explanation:

$\frac{dy}{dx}=x^{2}(y-1)\\\frac{1}{y-1} \text{ } dy=x^{2} \text{ } dx\\\int \frac{1}{y-1} \text{ } dy=\int x^{2} \text{ } dx\\\ln|y-1|=\frac{x^{3}}{3}+C\\$

From the initial condition,

$\ln|3-1|=\frac{0^{3}}{3}+C\\\ln 2=C$

So we have that $\ln |y-1|=\frac{x^{3}}{3}+\ln 2\\e^{\frac{x^{3}}{3}+\ln 2}=y-1\\2e^{\frac{x^{3}}{3}}=y-1\\y=2e^{\frac{x^{3}}{3}}+1$

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2 years ago

What % is 2000 from 40,000

3241004551 [841]

Answer:

5%

Step-by-step explanation:

2000/40000 = 0.05

All you do is divide both numbers and you'll get the decimal form which in this case is 0.05 which means 5%.

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3 years ago

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