All categories

2 years ago

Can somebody help me with this it a little hard

Mathematics

1 answer:

raketka [301]2 years ago

7 0

24 3-pound bags(72 pounds) and 10 5-pound bags(50 pounds). 72+50 = 122 pounds, 24+10 bags = 34 bags.

You might be interested in

Directions: Follow the instructions for the following inequalities. 1. 4<7 Multiply both sides by 7 , then by 6, then by 3, t

worty [1.4K]

In the given question, we have to perform this experiment and test if the inequality is true or not.

Adding the same value to both sides of the inequality will not change the inequality sign

<h3>Inequality</h3>

data;

4 < 7
11 > - 2
-4 < - 2
-8 < 8

In the first case, we would do follow the process

$6 * 7 * 4 < 7 * 7 * 6\\168 < 294$

In the second case;

$11 > - 2\\\\5 + 11 > -2 + 5\\3 + 16 > 3 + 3\\4+ 19 > 6 + 4\\23 > 10$

In all the scenarios, the operations did not affect the inequality sign because we are doing it equally to all side and at the end of the day nullifies each other.

Learn more on inequality here;

brainly.com/question/24372553

#SPJ1

4 0

1 year ago

Consider a data set containing the following values: 70 65 71 78 89 68 50 75 The mean of the preceding values is 70.75. The devi

Leokris [45]

Answer:

If this is the sample data, the sample variance is 125.07 and the sample standard deviation is 11.18

If this is a population data, the population variance is 109.44 and the population standard deviation is 10.46

Step-by-step explanation:

We are given the following data-set:

70, 65, 71, 78, 89, 68, 50, 75

Deviations from the mean

–0.75, –5.75 , 0.25, 7.25, 18.25, –2.75, –20.75, 4.2

Sample size, n = 8

Sample:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

Sum of squares of differences =

0.5625 + 33.0625 + 0.0625 + 52.5625 + 333.0625 + 7.5625 + 430.5625 + 18.0625 = 875.5

$s = \sqrt{\dfrac{875.5}{7}} = 11.18$

$\text{ Sample variance} = s^2 = 125.07$

Population:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

Sum of squares of differences =

0.5625 + 33.0625 + 0.0625 + 52.5625 + 333.0625 + 7.5625 + 430.5625 + 18.0625 = 875.5

$\sigma = \sqrt{\dfrac{875.5}{8}} = 10.46$

$\text{ Population variance} = \sigma^2 = 109.44$

7 0

2 years ago

Please help with this question. Thanks.

Kipish [7]

Answer:

FJH is an obtuse angle, so it is automatically larger than 90. Answer is D.

6 0

2 years ago

The speed limit is 25 miles per hour, but it seems that many of the cars are going faster. The city places 2 counter strips 360

klio [65]

You figure out how long it would take a car traveling at 25 mph
to cover 360 ft. Any driver who does it in less time is speeding.

   (25 mi/hr) · (5,280 ft/mile) · (1 hr / 3,600 sec)

   = (25 · 5280 / 3600) ft/sec = (36 and 2/3) feet per second.

To cover 360 ft at 25 mph, it would take

360 ft / (36 and 2/3 ft/sec) = 9.82 seconds .

Anybody who covers the 360 feet in less than 9.82 seconds
is moving faster than 25 mph.
_________________________________

If you're interested, here's how to do it in the other direction:

Let's say a car covers the 360 feet in ' S ' seconds.
What's the speed of the car ?

   (360 ft / S sec) · (1 mile / 5280 feet) · (3600 sec/hour)

   = (360 · 3600) / (S · 5280) mile/hour

   = 245.5 / S miles per hour .

The teacher timed one car crossing both strips in 7.0 seconds.
How fast was that car traveling ?

   245.5 / 7.0 = 35.1 miles per hour

Another teacher timed another car that took 9.82 seconds to cross
both strips. How fast was this car traveling ?

245.5 / 9.82 = 25 miles per hour

7 0

2 years ago

Help plz!! Has to do with concepts

aleksley [76]

Answer:

f(2) =4, f(4) =4, x=8, x=3

Step-by-step explanation:

f(2) is asking for what the y value is when x is 2. f(4) is also asking what the y value is when x=4. When y=-2 the x is 8 and when the y=6 x=4. I hope this is helpful.

3 0

1 year ago