Two mechanics worked on a car. The first mechanic worked for

Felicity set the thermostat of her refrigerator to 37°F. The refrigerator temperature t in degrees Fahrenheit h hours after the

kirill115 [55]

Answer:

Period=6

Step-by-step explanation:

Given:

t=1cos(1.05h)+37

Using acos(bx-c)+d to find the period of the given function

amplitude= a

period= 2π/Bb

phase shift=c (positive is to the left)

vertical shift=d

comparing with t=1cos(1.05h)+37, we get

a=1

b=1.05

c=0

d=37

period= 2π/b

=2π/1.05

=5.983

=6

Period of function is 6, after every 6 hours the refrigerator sensor will reach its maximum temperature and the cycle will move towards reducing temperature i.e it'll reach the minimum temperature then again the cycle will move upwards raising the temperature to maximum and so one period will be completed!

5 0

3 years ago

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Pls help will mark as brainliest

ASHA 777 [7]

Answer:

Step-by-step explanation:

8 0

3 years ago

Complete the square to form a perfect square trinomial x^2-14x+

fiasKO [112]

Answer: x^2 - 14x + 49

Explanation:

1) Divide the coefficient of x by 2:

14 / 2 = 7

2) so you have to add 7^2 = 49

x^2 - 14x + 49

3) that trinomial is equivalent to:

=> (x - 7)^2

4) prove that using the formula (a - b)^2 = a^2 - 2ab + b^2

(x - 7)^2 = x^2 - 14x + 49

Then you have to add 49 to complete the square. and form a perfect square trinomial.

7 0

3 years ago

The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let µ denote the true average re

Zolol [24]

Answer:

a) $df=n-1=16-1=15$

The statistic calculated is given by t=3.3

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

$p_v =P(t_{15}>3.3)=0.0024$

So since the p value is lower than the significance level $pv$ we reject the null hypothesis.

b) $df=n-1=8-1=7$

The statistic calculated is given by t=1.8

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

$p_v =P(t_{7}>1.8)=0.057$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

c) $df=n-1=26-1=25$

The statistic calculated is given by t=-0.6

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

$p_v =P(t_{25}>-0.6)=0.723$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation

$\bar X$ represent the sample mean

$s$ represent the standard deviation for the sample

$n$ sample size

$\mu_o =20$ 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 20, the system of hypothesis would be:

Null hypothesis: $\mu \leq 20$

Alternative hypothesis: $\mu > 20$

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".

(a) n = 16, t = 3.3, a = 0.05, P-value =

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

$df=n-1=16-1=15$

The statistic calculated is given by t=3.3

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

$p_v =P(t_{15}>3.3)=0.0024$

So since the p value is lower than the significance level $pv$ we reject the null hypothesis.

(b) n = 8, t = 1.8, a = 0.01, P-value =

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

$df=n-1=8-1=7$

The statistic calculated is given by t=1.8

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

$p_v =P(t_{7}>1.8)=0.057$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

(c) n = 26,t = -0.6, P-value =

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

$df=n-1=26-1=25$

The statistic calculated is given by t=-0.6

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

$p_v =P(t_{25}>-0.6)=0.723$

So since the p value is higher than the significance level $pv>\alpha$ we FAIL to reject the null hypothesis.

3 0

3 years ago

8 w + 12 = -12<br> Helpppppp

aliina [53]

Answer:

w = -3

Step-by-step explanation:

8w + 12 = -12

use inverse operations:

8w + 12 = -12

-12 -12

8w = -24

/8 /8

w = -3

7 0

3 years ago

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