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shusha [124]
3 years ago
10

you have a case of our mom's each one weighs 1.4 pounds and weight of 28 lb how many watermelons do you have

Mathematics
1 answer:
LenaWriter [7]3 years ago
3 0

Answer:

20 watermelons

Step-by-step explanation:

28/1.4 = 20

Plz give me brainiest!!!!

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Help me pls. I need help with the question
likoan [24]

Answer:

$173.90

Step-by-step explanation:

The most obvious way to answer this question is to multiply $9.40 by 18.5

This would result in 173.90.

But I realize nobody like to deal with decimals in a multiplication problem, so here's the way I would do it (<em>especially if you don't have a calculator!</em>)

The strategy is to break it down!

Multiply $9.40 x 18 hours to start, since this is easier.

9.4 x 18 = 169.2

So, that's $169.20 for 18 hours of work.

We aren't finished, because Abdel worked for 18.5 hours.

So, what half of $9.40?

9.4 / 2

= 4.7

-or-

$4.70

So, knowing that Abdel makes $4.70 in half an hour, lets just add that onto the $169.20 we got earlier.

169.2 + 4.7 = 173.9

-or-

$173.90 for 18.5 hours of work!

8 0
3 years ago
Suppose that resting pulse rates for healthy adults are found to follow a Normal distribution, with a mean of 69 beats per minut
laiz [17]

Answer:

Hi. im an online  tutor and i can assist you with your assignments. check out toplivewriters.com for support

Step-by-step explanation:

5 0
2 years ago
A road 4 inches long on a map is 80 miles long in real life. A river is 2 ½ inches long on the map and is 50 miles long in real
jonny [76]
Not Proportional. Sorry if it’s wrong
7 0
2 years ago
In order to ensure efficient usage of a server, it is necessary to estimate the mean number
juin [17]

Answer:

a. [36.19;39.21]

b. Reject the null hypothesis. The population mean of users that are connected at the same time is greater than 35.

Step-by-step explanation:

Hello!

Your study variable is,

X: "number of users of one server at a time"

The objective is to estimate the mean, for this, a sample of n=100 times was taken and the standard deviation S= 9.2 and the sample mean is X[bar]= 37.7 were calculated.

You need to study the population mean, for this you need your variable to have at least normal distribution. Since you don't have information about its distribution, but the sample is big enough (n≥30) you can apply the Central Limit Theorem and approximate the distribution of the sample mean X[bar] to normal:

X[bar]≈N(μ;σ²/n)

a. With this approximation, you can construct the 90% Confidence Interval using the approximate Z

[X[bar] ± Z_{1-\alpha /2} * S/√n]

Z_{1-\alpha /2} = Z_{0.95} = 1.64

[37.7± 1.64* 9.2/√100]

[36.19;39.21]

b. You need to test if the population mean is greater than 35 with a level of significance of 1%.

The hypothesis is:

H₀: μ ≤ 35

H₁: μ > 35

α: 0.01

This is a one-tailed test so you have only one critical level (right tail):

Z_{1\alpha } = Z_{0.99} = 2.33

This means that if the value of the calculated statistic is equal or greater than 2.33 you will reject the null Hypothesis.

If the value is less than 2.33 you will support the null hypothesis.

The statistic is:

Z=<u> X[bar] - μ </u>= <u> 37.7 - 35 </u> = 2.93

       S/√n           9.2/10

The value 2.93 > 2.33, so you reject the null hypothesis. This means that the population mean of users that are connected at the same time is greater than 35.

<u><em>Note: </em></u><em>To make the decision using the interval calculated on a), the hypothesis should have been two-tailed and the confidence and significance levels complementary.</em>

I hope it helps!

7 0
3 years ago
What is the 4th equivalent fraction to <br> 5/6
Lady bird [3.3K]

Answer:

20/24

Step-by-step explanation: Hope this helps

3 0
2 years ago
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