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2 years ago

A hot air balloon went from an elevation of 5,275 feet to an elevation of 3,128 feet in

Mathematics

1 answer:

Debora [2.8K]2 years ago

5 0

Answer:

0.942 ft/s

Step-by-step explanation:

The rate of descent is the rate at which the hot air balloon is descending, that is, how fast it is descending.

We calculate this by finding how far it has moved and then dividing it by the time it took it to descend.

It moved from 5275 ft to 3128 ft. The displacement is:

5275 - 3128 = 2147 ft

The time taken is 38 minutes. In seconds, it is:

38 * 60 = 2280 seconds.

Therefore, the rate of descent is:

2147 / 2280 = 0.942 ft/s

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Cynthia rounds a number, x,to one decimal place. the result is 6.3.

gladu [14]

The answer is: $6.25\leq6.3$

The explanation is shown below:

1. Cynthia rounds the number, which is identified as $x$ , to one decimal place and the result is 6.3.

2. Based on this, we know that $x$ could have been between 6.25 and 6.35. Therefore, the error interval for $x$ is given by:

$6.25\leq6.3$

Where $\leq$ indicates that the value 6.25 is included and indicates that the value 6.35 is not included (Because if $x$ had been exactly 6.35, Cynthia would round up to 6.4).

7 0

2 years ago

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

2 years ago

Michaela’s class used a spinner that has 6 spaces of equal area in which the arrow can land. The areas are numbered 1, 2, 3, 4,

diamong [38]

The number of times the spinner landed on a space numbered greater than 4 = 167

Step-by-step explanation:

Step 1 :

Given,

Number of equal area spaces in the spinner = 6

Number of times the spinner was spun = 500

We need to find the number of times the spinner landed on a space numbered greater than 4

Step 2 :

Total number of outcome = 6

Favorable outcomes are = Numbers greater than 4 = 5,6

Total number of favorable outcome = 2

Therefore Probability that the spinner will land on a number that is greater than 4 is $\frac{2}{6}$ = $\frac{1}{3}$

The number of times that the spinner landed on a space numbered greater than 4 = 500 × $\frac{1}{3}$ = 166.67 = 167 (rounded off to the nearest integer)

Step 3 :

Answer :

The number of times the spinner landed on a space numbered greater than 4 = 167

4 0

2 years ago

Kathy has $50,000 to invest today and would like to determine whether it is realistic for her to achieve her goal of buying a ho

olga_2 [115]

Answer:

a) About 12%

Step-by-step explanation:

We need to find the interest rate required to achieve her goal, so we will need to use the interest-compound formula:

$FV=PV(1+i)^{n}$

Where:

PV= Present Value

i= interest rate

FV= Future Value

n= number of periods

replacing the data provided:

$150.000=50.000(1+i)^{10}$

solving for i:

first, divide both sides by 50.000 to simplify the equation:

$3=(1+i)^{10}$

Take $10^{th}$ roots of both sides:

$1+i=$ ± $\sqrt[10]{3}$

solve for i:

$i=$ ± $\sqrt[10]{3} -1$

We get two answers, but we look for a coherent value. So we take the positive one:

$i=11.6123174$ ≈12

6 0

2 years ago

Mixing two kinds of candy the price of which was $2 and $4 per pound, Ella got a 10-lb mix of candy, which cost $2.90 per pound.

topjm [15]

1. You have the following information given in the problem above:

- Ella mixed<span> two kinds of candy the price of which was $2 and $4 per pound.
- Ella got a 10-lb mix of candy, which cost $2.90 per pound.
</span>
2. Therefore, let's call:

x: pounds of the first kind of candy.
y: pounds of the second kind of candy.

3. Then, you have:

2x+4(10-x)=(2.90)(10)

4. When you clear x, you obtain:

2x+40-4x=29
-2x=29-40
x=-11/-2
x=5.5 pounds

x+y=10
y=10-x
y=10-5.5
y=4.5 pounds

8 0

2 years ago