All categories

2 years ago

I will give brand list and hurry pleas

Mathematics

1 answer:

Artyom0805 [142]2 years ago

6 0

Answer:

3/2 cm/s

Step-by-step explanation:

speed = gradient of the line (slope)

To calculate the slope divide the change in y by the change in x:

(9 - 0) / (6 - 0) = 9/6 = 3/2 cm/s

You might be interested in

Confused from where I left off PLEASE help

Leviafan [203]

Not sure the right equation so i did both....

#1 - √12m√15m=1m√2

Simplifies to:

13.416408m2=1.414214m

Let's solve your equation step-by-step.

13.416408m2=1.414214m

Step 1: Subtract 1.414214m from both sides.

13.416408m2−1.414214m=1.414214m−1.414214m

13.416408m2−1.414214m=0

Step 2: Factor left side of equation.

m(13.416408m−1.414214)=0

Step 3: Set factors equal to 0.

m=0 or 13.416408m−1.414214=0

m=0 or m=0.105409

Answer: m=0 or m=0.105409

#2 - √12m√15m

Simplifies to:

13.416408m2

=13.416408*m2

=13.416408*(m*m)

=13.416408m2

Can I get a brainliest Please & Thank you...

4 0

3 years ago

Read 2 more answers

5x - 2z = 8 -2y + 9z = 50 6x + 9y + 9z = 96

sergiy2304 [10]

Answer:

there r algebra calulators you could use

Step-by-step explanation:

5 0

3 years ago

Calculate 21% of $6.85 = $

choli [55]

Answer: $1.44

Step-by-step explanation: hope this helps please give me brainliest.

6 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

Which expression shows a sum of five terms? (3 points)

Sav [38]

Answer:

5 + x + 5 + x + x

Step-by-step explanation:

We are given 4 expressions and we have to determine which expressions shows a sum of five terms.

We use the process of elimination:

5(x + 5)

=5x+25 (distributing 5)

This contains sum of only two terms, eliminate.

It contains only 1 term, eliminate.

5 + x + 5 + x + x

It contains 5 terms 5,x,5,x and x, KEEP!

Only contains 1 term, eliminate.

With this said, were left with:

5 + x + 5 + x + x

Which is the correct answer! :)

6 0

2 years ago