Which number is irrational? A.-4 B.2/9 C.sqrt of 11 D.8.26... (repeating 6 only)

4. A typical smartphone battery (yes that includes iphones) degrades at approximately 1%

Zielflug [23.3K]

Answer:

a. $f(t) = (0.99)^{t} 100%\\$

b. The x - intercept represents the time at which the battery life drops to 0 %.

c. The y - intercept represents the initial battery life.

Step-by-step explanation:

a. Construct an equation using f(t) and t that represents the relationship between

"f(t)" the battery capacity of a typical smartphone after "t" months and "L" the

number of months the smartphone has been in normal use.

Let the initial battery life be I.

After one month, it decreases by 1%. So, our new value, I' = initial value - decrease = I - 1%I = 99%I.

After two months, it decreases by 1% to our new value I". So, our new value, I" = initial value - decrease = I' - 1%I' = 99%I' = 99%(99%)I = (99%)²I

After three months, it decreases by 1% to our new value I'". So, our new value, I"' = initial value - decrease = I" - 1%I" = 99%I" = 99%(99%)²I = (99%)³I

We see a pattern here.

After t months, f(t)

$f(t) = (0.99)^{t} 100%\\$

b. What does the horizontal intercept represent in context? (The x-intercept)

The x - intercept represents the time at which the battery life drops to 0 %.

c. What does the vertical intercept represent in context? (The y-intercept)

The y - intercept represents the initial battery life.

5 0

3 years ago

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found

joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the population mean, $\mu$ is ;

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

80% confidence interval for $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

98% confidence interval for $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

6 0

3 years ago

Can anybody help me please

Sliva [168]

Hello, your answer would be (-5, 5)

3 0

3 years ago

An electric motor uses 6kw at an efficiency of 63%. How much power does it deliver?

r-ruslan [8.4K]