Solve for y. y – (–19) = 25 a. –44 b.

Y=4x(8-x). Linear or nonlinear? explain.

NemiM [27]

Answer:

not linear

Step-by-step explanation:

this equation becomes y = -4x² + 32x which is a quadratic equation that, when graphed, forms an upside down u-shaped curve (upside down because the x² term is negative; if it's positive then the curve is u-shaped)

4 0

3 years ago

Read 2 more answers

Split the number 76 into the ratio 6:3:7:3 .

kompoz [17]

Answer:

24:12:28:12

Step-by-step explanation:

76÷19=4

6×4:3×4:7×4:3×4

=24:12:28:12

3 0

3 years ago

A yacht is anchored 90 feet offshore

Stells [14]

The height of the lighthouse is 44 feet.

Step-by-step explanation:

You are given more information than you need.

Use the right triangle concept,

A yacht is anchored 90 feet offshore from the base of a lighthouse.

Base = 90 feet
Height = h feet

The angle of elevation from the boat to the top of the lighthouse is 26 degrees.

Using one side length and the angle only:

The trigonometric formula is given as,

tan = height / base

tan[26º] = h ⁄ 90

h = 90 × tan[26º]

h = 43.9 ft

h = 44 feet

Therefore, the height of the lighthouse is 44 feet.

8 0

3 years ago

Read 2 more answers

It is claimed that automobiles are driven on average more than 20,000 kilometers per year. To test this claim, 100 randomly sele

kow [346]

Answer:

$t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974$

$p_v =P(t_{99}>8.974)=9.43x10^{-15}$

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.

Step-by-step explanation:

1) Data given and notation

$\bar X=23500$ represent the sample mean

$s=3900$ represent the sample standard deviation

$n=100$ sample size

$\mu_o =2000$ 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.

We need to conduct a hypothesis in order to determine if the true mean is higher than 20000, the system of hypothesis would be:

Null hypothesis: $\mu \leq 2000$

Alternative hypothesis: $\mu > 2000$

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".

Calculate the statistic

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

$t=\frac{23500-20000}{\frac{3900}{\sqrt{100}}}=8.974$

Calculate the P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=100-1=99$

Since is a one-side upper test the p value would be:

$p_v =P(t_{99}>8.974)=9.43x10^{-15}$

Conclusion

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we to reject the null hypothesis, and the actual mean is significantly higher than 20000.

8 0

3 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

3 years ago

Solve for y. y – (–19) = 25 a. –44 b. –6 c. 6 d. 44