2 whole and 2/4 divided by 4 wholes and 1/8

Find two consecutive odd integers such that 62 more than the lesser is five times the greater.

lidiya [134]

Idk really know srry

8 0

2 years ago

Read 2 more answers

25 Pts + Brainliest for the first correct answer.

NeTakaya

6.)Corresponding

7.)Alternate Interior

8.)Vertical

9.)Alternate Exterior

10.)Adjacent

8 0

3 years ago

35 is what percent of 150? Enter your answer in decimal form to the nearest hundreth.

Sophie [7]

Answer:

23.33%

Step-by-step explanation:

Solution for 35 is what percent of 150:

35: 150*100 =

(35*100): 150 =

3500: 150 = 23.33

Now we have: 35 is what percent of 150 = 23.33

Question: 35 is what percent of 150?

Percentage solution with steps:

Step 1: We make the assumption that 150 is 100% since it is our output value.

Step 2: We next represent the value we seek with x.

Step 3: From step 1, it follows that 100%=150.

Step 4: In the same vein, x%=35.

Step 5: This gives us a pair of simple equations:

100%=150(1).

x%=35(2).

Step 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS

(left hand side) of both equations have the same unit (%); we have

\frac{100\%}{x\%}=\frac{150}{35}

Step 7: Taking the inverse (or reciprocal) of both sides yields

\frac{x\%}{100\%}=\frac{35}{150}

Therefore, 35 is 23.33% of 150.

7 0

2 years ago

Y=12x-17 Slope what is it

oee [108]

Answer: -204

Step-by-step explanation: 12 X -17 +Y is the same thing but you can say it this way and its easier to understand but it equals -204/(negative 204) Hope this helps you! :3

6 0

3 years ago

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You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confide

salantis [7]

Answer:

With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

Step-by-step explanation:

We are given that a random sample of 60 home theater systems has a mean price of$131.00. Assume the population standard deviation is$18.80.

Firstly, the pivotal quantity for 90% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean price = $131

$\sigma$ = population standard deviation = $18.80

n = sample of home theater = 60

$\mu$ = population mean

Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $131-1.645 \times {\frac{18.8}{\sqrt{60} } }$ , $131+1.645 \times {\frac{18.8}{\sqrt{60} } }$ ]

= [127.01 , 134.99]

Therefore, 90% confidence interval for the population mean is [127.01 , 134.99].

Now, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean price = $131

$\sigma$ = population standard deviation = $18.80

n = sample of home theater = 60

$\mu$ = population mean

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.

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

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $131-1.96 \times {\frac{18.8}{\sqrt{60} } }$ , $131+1.96 \times {\frac{18.8}{\sqrt{60} } }$ ]

= [126.24 , 135.76]

Therefore, 95% confidence interval for the population mean is [126.24 , 135.76].

Now, with 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

7 0

3 years ago