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3 years ago

Answer PLEASE doing brainliest

Mathematics

2 answers:

Maurinko [17]3 years ago

6 0

Answer:

Step-by-step explanation:

First start by simplifying each expression.

To simplify $\frac{4}{5}$ ÷ $\frac{1}{12}$ you can use the method: Keep Change Flip to get $\frac{4}{5}$ × $\frac{12}{1}$ . Now you can simplify to get $\frac{48}{5}$ . Then divide to get the decimal form of 9.6

For the expression: $1\frac{9}{13}$ ÷ $\frac{1}{8}$ . First you must get both into improper fractions: $\frac{22}{13}$ ÷ $\frac{1}{8}$ . Now you can Keep Change Flip: $\frac{22}{13}$ × $\frac{8}{1}$ . Now multiply: $\frac{176}{13}$ . Now you can divide to get the decimal form of about 13.54

Repeat the steps for the next expression. $1\frac{9}{13}$ ÷ $\frac{1}{12}$ . Reduce both fractions to improper fractions: $\frac{22}{13}$ ÷ $\frac{1}{12}$ . Not Keep Change Flip: $\frac{22}{13}$ × $\frac{12}{1}$ . Now multiply: $\frac{264}{13}$ . Next divide to get the decimal form of about 20.31.

For the last expression: $\frac{4}{5}$ ÷ $\frac{1}{8}$ . First start by Keep Change Flip: $\frac{4}{5}$ × $\frac{8}{1}$ . Now multiply to get: $\frac{32}{5}$ . Lastly divide to get the decimal form of 6.4.

Now that you have all decimal forms, you can now put them in order:

1. Least - $\frac{4}{5}$ ÷ $\frac{1}{8}$

2. $\frac{4}{5}$ ÷ $\frac{1}{12}$

3. $1\frac{9}{13}$ ÷ $\frac{1}{8}$

4. Greatest - $1\frac{9}{13}$ ÷ $\frac{1}{12}$

user100 [1]3 years ago

5 0

Answer:

least

1. 4/5÷1/8

2. 4/5÷1/12

3. 1 19/13÷18

4. 1 9/13÷1/12

greatest

answers

1. - 32/5 (6.4, in decimal form)

2. - 48/5 (9.6, in decimal form)

3. - 176/13 (13.53846154, in decimal form)

4. - 264/13 (20.30769231, in decimal form)

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Helppp me plsssssssss<br><br>

Oliga [24]

Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

$\sf \:\:\:\:\:\:\:\:\:\:x_{k}=35$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k}=50$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42$
$\sf \:\:\:\:\:\:\:\:\:\:h=5$

${\bf \:\: {By\:using\:the\: formula}} \\ \\$

$\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\$

$\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\$

${\large{\frak{\pmb{\underline{Additional\: information }}}}}$

MODE

Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

In a frequency distribution the class having maximum frequency is called the modal class.

${\bf{\underline{Formula\:for\: calculating\:mode:}}} \\$

${\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\$

Where,

$\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.$

$\small \blue{ \bigstar}$ $\sf \: f_{k}=frequency\:of\:the\:modal\:class$

$\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class$

$\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class$

$\small \purple{ \bigstar}$ $\sf \: h= width \:of\:the\:class\:interval$

7 0

3 years ago

6% of $120 find percent<br> 0.5% of $44 find percent

just olya [345]

.06*120=$7.20
.005*44=$0.22

6 0

3 years ago

Read 2 more answers

To estimate the mean height μ of male students on your campus,you will measure an SRS of students. You know from government data

nexus9112 [7]

Answer:

a) $\sigma = 0.167$

b) We need a sample of at least 282 young men.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

This Zscore is how many standard deviations the value of the measure X is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

(a) What standard deviation must x have so that 99.7% of allsamples give an x within one-half inch of μ?

To solve this problem, we use the 68-95-99.7 rule. This rule states that:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviations of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we want 99.7% of all samples give X within one-half inch of $\mu$ . So $X - \mu = 0.5$ must have $Z = 3$ and $X - \mu = -0.5$ must have $Z = -3$ .

$Z = \frac{X - \mu}{\sigma}$

$3 = \frac{0.5}{\sigma}$

$3\sigma = 0.5$

$\sigma = \frac{0.5}{3}$

$\sigma = 0.167$

(b) How large an SRS do you need to reduce the standard deviationof x to the value you found in part (a)?

You know from government data that heights of young men are approximately Normal with standard deviation about 2.8 inches. This means that $\sigma = 2.8$