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3 years ago

Isaiah has $145 in his savings account. He earns $36 a week mowing lawns.

Mathematics

1 answer:

valentina_108 [34]3 years ago

3 0

Answer:

8 weeks

Step-by-step explanation:

Amount saved from mowing lawns = $433 - $145

= $288

Number of weeks to save $288 from $36 a week = 288/36 = 8 weeks

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NeTakaya

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3 years ago

Jeff hikes 1/2 mile every 15 minutes, or 1/4 hour. Lisa hikes 1/3 mile every 10 minutes, or 1/6 hour. How far do they each hike

inessss [21]

Answer:

they each hike 2 miles

Step-by-step explanation:

Jeff: 15=1/2

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3 years ago

Sam the owl is looking down at a 24° angle from the top of a tree that is 10 ft tall, when he spots a bird on the ground. How fa

Liula [17]

Answer:

24.59 feet

Step-by-step explanation:

Let x represent the distance between Sam and the bird.

We have been given that am the owl is looking down at a 24° angle from the top of a tree that is 10 ft tall, when he spots a bird on the ground. We are asked to find the distance between Sam and the bird.

We can see from our attachment that Sam, the bird and angle of depression forms a right triangle with respect to ground. We can see that side 10 ft is opposite side and and side x is hypotenuse for the 24 degree angle.

$\text{sin}=\frac{\text{Opposite}}{\text{Hypotenuse}}$

$\text{sin}(24^{\circ})=\frac{10}{x}$

$x=\frac{10}{\text{sin}(24^{\circ})}$

$x=\frac{10}{0.406736643076}$

$x=24.5859333557$

$x\approx 24.59$

Therefore, the bird is approximately 24.59 feet away from Sam.

8 0

3 years ago

The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is: