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amid [387]
4 years ago
9

Find the range of 88, 16, 34, 48 and 27

Mathematics
2 answers:
nikitadnepr [17]4 years ago
7 0
Hi Lea.

The range is the difference between the largest number and the smallest number. Our largest number is 88, and our smallest number is 16. To find the range, we need to subtract 16 from 88.

88 - 16 = 72

72 is your range. 

I hope this helps!
frozen [14]4 years ago
3 0
The range is 72. hope this helps! (subtract the lowest number from the highest number in the set)
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A jar holds no more than 500 buttons. If 5 people bring 330 buttons each, how many jars are necessary to hold all of the buttons
Strike441 [17]

Answer:

We know

1 jar = 500 buttons

5 people bring 330 buttons EACH

We can find/solve by...

people x buttons = total buttons

total buttons / 500 = number of jars needed

So..

5 x 330 = 1650

1650 / 500 = 3.3

Your answer:

4 jars

Explanation

It's impossible to have 3.3 jars, so you must round up.

Step-by-step explanation:

8 0
3 years ago
Find the product.<br> a.<br> 5d - 79<br> b. 12 ab. 3cd
Solnce55 [7]

Answer:

35dg & 36abcd

Step-by-step explanation:

Multiply 7 times 5 then add the remaining terms for equation 1. For equation 2 multiply 12 by 3 for 36.

6 0
3 years ago
A group of students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are list
Kitty [74]

Answer:

It does not appear ​that, as a​ group, the students are reasonably good at estimating one minute.

Step-by-step explanation:

We are given the following data in the question:

75, 88, 51, 73, 49, 31, 69, 74, 72, 59, 72, 81, 99, 101, 73

Formula:

\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.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{1067}{15} = 71.13

Sum of squares of differences = 4739.733

S.D = \sqrt{\frac{4739.733}{14}} = 18.39

Population mean, μ = 60 minutes

Sample mean, \bar{x} = 71.13 minutes

Sample size, n = 15

Alpha, α = 0.10

Sample standard deviation, s = 18.39 minutes

First, we design the null and the alternate hypothesis

H_{0}: \mu = 60\text{ minutes}\\H_A: \mu \neq 60\text{ minutes}

We use Two-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{71.13 - 60}{\frac{18.39}{\sqrt{15}} } = 2.34

Calculating the p-value from the table, we have,

P-value = 0.034354

Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Thus, we conclude that it does not appear ​that, as a​ group, the students are reasonably good at estimating one minute.

7 0
3 years ago
there are 19 pieces of candy in the candy jar Mrs Breen lost her mind and started eating all these mindlessly she realize she ha
kkurt [141]
The percent of candy she ate is 42%
4 0
3 years ago
How do I solve for x​
Free_Kalibri [48]

Answer:

Can you explain a lil bit

Step-by-step explanation:

Dont understand by that

4 0
3 years ago
Read 2 more answers
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