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3 years ago

What is the range of the data? 53, 47, 60, 45, 62, 59, 65, 50, 56, 48

Mathematics

2 answers:

Nataly [62]3 years ago

8 0

Answer:

12 hope it helps

Step-by-step explanation:

dolphi86 [110]3 years ago

6 0

Answer:20

Step-by-step explanation: To find the range of a set of data,

1. find your largest data point, in this case 65

2.find you lowest data point, 45

3. subtract the lowest number from the highest 65-45=20

4 this is your range

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Select all of the situations which can be represented by the expression 25.75x + 10.

mafiozo [28]

Answer: 1. no

2. yes

3. no

4. yes

Step-by-step explanation:

8 0

3 years ago

Zack had 20 people show up for his first exercise class he had 36 come to second class what is the percent increased

Mariulka [41]

The percent increase is 80%

Step-by-step explanation:

Percent increase is given by:

$Percent\ Increase=\frac{Increase}{Previous\ value}*100\\While\\Increase= New\ value - Old\ value\\$

Here

New value = 36

Old Value = 20

Increase = 36-20 = 16

So,

$Percent\ increase=\frac{16}{20}*100\\=0.8*100\\=80\%$

Hence,

The percent increase is 80%

Keywords: Percentage, Percent

Learn more about percentage at:

#LearnwithBrainly

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3 years ago

Analyze the graph below to identify the key features of the logarithmic function.

dmitriy555 [2]

Answer:

is there just 2 answers too this question? anyways the answer b

4 0

3 years ago

Evaluate 8w - 2p if w = 6 and p = -5

zysi [14]

Known facts

w = 6
p = -5

8w - 2p = 8(6) - 2(-5) = 48 + 10 = 58

Answer: 58

Hope that helps!

4 0

2 years ago

Read 2 more answers

A study was recently conducted to estimate the mean cholesterol for adult males over the age of 55 years. From a random sample o

Komok [63]

Answer: Correct critical value = 2.2622

Step-by-step explanation:

Confidence interval for population mean when population standard deviation is unknown:

$\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$ , where $\overline{x}$ = sample mean, n= sample size, s= sample standard deviation, $t_{\alpha/2}$ = two-tailed t value.

As per given: n= 10

degree of freedom : df = n-1=9

$\overline{x}=242.6\\\\s=73.33\\\\\alpha=5\%=0.05$

Critical t-value : $t_{\alpha/2,df}=t_{0.05/2,9}=t_{0.025,9}=2.2622$

So, the 95 percent confidence interval estimate for the mean :

$242.6\pm (2.2622)\dfrac{73.33}{\sqrt{10}}\\\\=242.6\pm (2.2622)(23.19)\\\\\approx242.6\pm52.46\\\\=(242.6-52.46,\ 242.6+52.46)\\\\=(190.14,\ 295.06)$

The 95 percent confidence interval estimate for the mean:(190.14, 295.06)

8 0

3 years ago