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2 years ago

If a car can go 204 miles in 3 7/8 hours, how far can it go in 2 1/5 hours

Mathematics

1 answer:

Naya [18.7K]2 years ago

7 0

Answer:

115.5

Step-by-step explanation:

the car is traveling at precisely 52.5 miles an hour

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Divide. 413÷129 Enter your answer, as a mixed number in simplest form, in the box.

kompoz [17]

For this case we must divide the following expression:

$\frac {413} {129}$

The expression can not be simplified. When dividing we have that its decimal form is given by:

3,201550

If we convert to a mixed number we have:

$3 \frac {26} {129}$

Verifying:

$3 \frac {26} {129} = \frac {129 * 3 + 26} {129} = \frac {413} {129}$

Answer:

$3 \frac {26} {129}$

5 0

3 years ago

How might snakes running out of food affect other organisms such as hawks? please help

marissa [1.9K]

Answer:

The population of Hawks and other species would likely start to edge off.

Step-by-step explanation:

Commonly, a Food Chain is dependent on all the parts of it sataying at a perfect point. If the food that the Snakes eat starts to run out, more of them will die off. With less Snakes, the Hawks will then have less food, and that continues throughout the chain. The same thing would happen if there was too much food, except instead of the populations of species dying off, the Snake population would spike.

3 0

2 years ago

Hallie bought 20 lollipops for $0.25 each and 15 peppermint patties for $0.35. How much money did Hallie spend on candy?

Nastasia [14]

Answer:

She spent $10.25 on candy.

Step-by-step explanation:

Step 1- Multiply

20×0.25= 5

15×0.35= 5.25

Step 2- Add

5+5.25= 10.25

7 0

3 years ago

Read 2 more answers

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation:

Vikentia [17]

Answer:

$\= x = 18.5$ , $\sigma = 5.15$

$15.505 < \mu < 21.495$

$14.93 < \mu < 22.069$

Step-by-step explanation:

From the question we are are told that

The sample data is 21, 14, 13, 24, 17, 22, 25, 12

The sample size is n = 8

Generally the ample mean is evaluated as

$\= x = \frac{\sum x }{n}$

$\= x = \frac{ 21 + 14 + 13 + 24 + 17 + 22+ 25 + 12 }{8}$

$\= x = 18.5$

Generally the standard deviation is mathematically evaluated as

$\sigma = \sqrt{\frac{\sum (x- \=x )^2}{n}}$

$\sigma = \sqrt{\frac{\sum ((21 - 18.5)^2 + (14-18.5)^2+ (13-18.5)^2+ (24-18.5)^2+ (17-18.5)^2+ (22-18.5)^2+ (25-18.5)^2+ (12 -18.5)^2 )}{8}}$

$\sigma = 5.15$

considering part b

Given that the confidence level is 90% then the significance level is evaluated as

$\alpha = 100-90$

$\alpha = 10\%$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table the value is

$Z_{\frac{ \alpha }{2} } = 1.645$

The margin of error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E =1.645 * \frac{5.15 }{\sqrt{8} }$

=> $E = 2.995$

The 90% confidence interval is evaluated as

$\= x - E < \mu < \= x + E$

substituting values

$18.5 - 2.995 < \mu < 18.5 + 2.995$

$15.505 < \mu < 21.495$

considering part c

Given that the confidence level is 95% then the significance level is evaluated as

$\alpha = 100-95$

$\alpha = 5\%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E =1.96 * \frac{5.15 }{\sqrt{8} }$

=> $E = 3.569$

The 95% confidence interval is evaluated as

$\= x - E < \mu < \= x + E$

substituting values

$18.5 - 3.569 < \mu < 18.5 + 3.569$

$14.93 < \mu < 22.069$

8 0

3 years ago

Artemon [7]

Answer:

In Martian Poker, the game is played with 8 cards drawn from a 52 card deck.

What is the probability of a Martian drawing “two trips and a pair". They get 2

three of a kinds and a pair.

8 0

3 years ago