Write an inequality for the word phrase "x increased by 5 is at least 13"

Total money spent is $11425 buying 75 animals
each chicken costs $65 so 50 costed the farmer $3250
each cow costs $45 so 15 cows costed the farmer $675
ducks purchased were 10
to find how much the ducks costs we will add the money spent the farmer and subtract it from the total amount spent.
11425-3250-675=$7500
10 ducks = $7500
so one duck costs 7500/10=$750
the price for each duck is $750

4 0

3 years ago

Which is the graph of f (x) = 4 (one-half)^x?

grigory [225]

Answer:

answer is B. :)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

For a test of population proportion H0: p = 0.50, the z test statistic equals 1.05. Use 3 decimal places. (a) What is the p-valu

sergiy2304 [10]

Answer:

(a) The p-value of the test statistic is 0.147.

(b) The p-value of the test statistic is 0.294.

(c) The p-value of the test statistic is 0.8531.

(d) None of the p-values give strong evidence against the null hypothesis.

Step-by-step explanation:

The p-value is well defined as the probability,[under the null hypothesis (H₀)], of attaining a result equivalent to or greater than what was the truly observed value of the test statistic.

We reject a hypothesis if the p-value of a statistic is lower than the level of significance α.

The null hypothesis for the test of population proportion is defined as:

H₀: p = 0.50

The value of z-test statistic is,

z = 1.05

(a)

The alternate hypothesis is defined as:

Hₐ: p > 0.50

Compute the p-value of the test statistic as follows:

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

*Use a z-table for the probability value.

Thus, the p-value of the test statistic is 0.147.

(b)

The alternate hypothesis is defined as:

Hₐ: p ≠ 0.50

Compute the p-value of the test statistic as follows:

$p-value=2\times P(Z>1.05)\\=2\times 0.1469\\=0.2938\\\approx 0.294$

*Use a z-table for the probability value.

Thus, the p-value of the test statistic is 0.294.

(c)

The alternate hypothesis is defined as:

Hₐ: p < 0.50

Compute the p-value of the test statistic as follows:

$p-value= P(Z1.05)=1- 0.1469\\=0.8531$

*Use a z-table for the probability value.

Thus, the p-value of the test statistic is 0.8531.

(d)

The decision rule of the test is:

If the p-value of the test is less than the significance level α, then the null hypothesis is rejected at α% level of significance.

And if the p-value of the test is more than the significance level α, then the null hypothesis is failed to be rejected.

The most commonly used level of significance are:

α = 0.01, 0.05 and 0.10

The p-value for all the three alternate hypothesis are:

p-values = 0.147, 0.294 and 0.8531.

All the p-values are quite large compared to the α values.

Thus, none of the p-values give strong evidence against the null hypothesis.

The null hypothesis was failed to be rejected.

5 0

3 years ago