J____K_______L JK is x+3 KL is 2x-5 JL is 19 find the value of JK

Jada rode her bike at a constant speed for 3 hours. During that time, she traveled 7.2 miles. What was Jada's riding speed?

NeTakaya

Answer: 2.4 miles per hour

Step-by-step explanation: just do 7.2/3=2.4

or another way you could do this is do 2.4 x 3=7.2

3 0

3 years ago

Set of data with 350,000 numbers is normally

Rudiy27

Step-by-step explanation:

given a normal distribution with the given parameters the probability (= the % of the area of the distribution curve) for a number to be between 203 and 1803 is

0.9987

so, 99.87% of all numbers are expected to be in that range.

for 350,000 numbers that means

350,000×0.9987 = 349,545 numbers are expected to be between 203 and 1803.

7 0

2 years ago

Select the correct answer.

skad [1K]

Answer:

D

Step-by-step explanation:

7 is it because it is positive

6 0

3 years ago

An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concen

devlian [24]

Answer:

(a) A 95% confidence interval for the population mean is [433.36 , 448.64].

(b) A 95% upper confidence bound for the population mean is 448.64.

Step-by-step explanation:

We are given that article contained the following observations on degrees of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

420, 425, 427, 427, 432, 433, 434, 437, 439, 446, 447, 448, 453, 454, 465, 469.

Firstly, the pivotal quantity for finding the 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 = $\frac{\sum X}{n}$ = 441

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 14.34

n = sample size = 16

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X -2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X +2.131 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $441-2.131 \times {\frac{14.34}{\sqrt{16} } }$ , $441+2.131 \times {\frac{14.34}{\sqrt{16} } }$ ]

= [433.36 , 448.64]

(a) Therefore, a 95% confidence interval for the population mean is [433.36 , 448.64].

The interpretation of the above interval is that we are 95% confident that the population mean will lie between 433.36 and 448.64.

(b) A 95% upper confidence bound for the population mean is 448.64 which means that we are 95% confident that the population mean will not be more than 448.64.

6 0

3 years ago

Are the given ratios equivalent Explan why or why not 3 : 5 12 : 20

sasho [114]

Yes you take 3 times 4 and receive 12 and 5 times 4 and receive 20

4 0

3 years ago