In which quadrant would point (-4.5, -4.5) be located

What is 3.504 estimate to

tatuchka [14]

3.504 estimates to 4

5 0

3 years ago

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Which function has a range of {yly < 5}?

Alexeev081 [22]

Answer:

f(x) = -(x-4)^2 + 5

Step-by-step explanation:

The function $f(x) = -(x-4)^2 + 5$ is a quadratic function. Its graph looks like a parabola. The graph has a vertex of (4,5) and opens up downward.

Proving that $f(x) = -(x-4)^2 + 5 \le 5$ for all x:

$x^2 \ge 0 \Rightarrow (x-4)^2 \ge 0 \Rightarrow -(x-4)^2 \le 0 \Rightarrow -(x-4)^2 + 5 \le 5.$

8 0

3 years ago

The quotient of 24 and -8

rusak2 [61]

The answer to your question is -3

5 0

2 years ago

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A particular fruit's weights are normally distributed, with a mean of 405 grams and a standard deviation of 19 grams. If a fruit

cluponka [151]

Answer:

413.126

Step-by-step explanation:

Given a normal distribution :

Mean (m) = 405

Standard deviation (s) = 19

Weight of fruit will be greater than how many grams (x) if picked 17% of the time:

P(Z > x) = 0.17 ;

The Zscore for P(Z > x) = 0.17 equals 0.954 (Z probability calculator)

Zscore = (x - m) / s

0.954 = (x - 405) / 19

0.954 * 19 = x - 405

18.126 = x - 405

x = 18.126 + 405

x = 423.126

Weight will be greater than 423.126

8 0

3 years ago

1).Calculate T, when sample mean is 120, population mean is 100, standard deviation is 20 and smaple size is 10 using levels of

Sonbull [250]

Answer:

$t=\frac{120-100}{\frac{20}{\sqrt{10}}}=3.16$

$p_v =2*P(t_{9}>3.16)=0.012$

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we can reject the null hypothesis, and the true mean is significantly different from 100 at 5% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=120$ represent the sample mean

$s=20$ represent the standard deviation for the sample

$n=10$ sample size

$\mu_o =100$ represent the value that we want to test

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

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the mean is different from 100, the system of hypothesis would be:

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

Compute the test statistic

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-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".

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

$t=\frac{120-100}{\frac{20}{\sqrt{10}}}=3.16$

Now we need to find the degrees of freedom for the t distirbution given by:

$df=n-1=10-1=9$

Conclusion

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

$p_v =2*P(t_{9}>3.16)=0.012$

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we can reject the null hypothesis, and the true mean is significantly different from 100 at 5% of significance.

8 0

3 years ago

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