Find the lower quartile (first quartile) value of the data set below.

A circular flower bed is 16 m in diameter and has a circular sidewalk around it that is 2 m wide. Find the area of the sidewalk

Katyanochek1 [597]

59.69 is probably the answer. let me know.

7 0

3 years ago

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Which two number have an absolute value of 2

madam [21]

Answer:2 -2

Step-by-step explanation:absolute value is distance from zero so both of those are the same distance from zero

3 0

3 years ago

So,some please help me with this question !

Tamiku [17]

$\qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star$

$\qquad❖ \: \sf \: \angle C = 63 \degree$

$\textsf{ \underline{\underline{Steps to solve the problem} }:}$

$\qquad❖ \: \sf \:x + 4x + 12 + x=180°$

( Angle EBA= Angle DBC, and the three angles sum upto 180° due to linear pair property )

$\qquad❖ \: \sf \:6x + 12 = 180$

$\qquad❖ \: \sf \:6x = 180 - 12$

$\qquad❖ \: \sf \:6x = 168$

$\qquad❖ \: \sf \:x = 28 \degree$

Next,

$\qquad❖ \: \sf \: \angle C + \angle D + x = 180°$

$\qquad❖ \: \sf \: \angle C +3x + 5 + x = 180°$

$\qquad❖ \: \sf \: \angle C +4x = 180 - 5$

$\qquad❖ \: \sf \: \angle C +4(28) = 175$

( x = 28° )

$\qquad❖ \: \sf \: \angle C +112 = 175$

$\qquad❖ \: \sf \: \angle C = 175 - 112$

$\qquad❖ \: \sf \: \angle C = 63 \degree$

$\qquad \large \sf {Conclusion} :$

$\qquad❖ \: \sf \: \angle C = 63 \degree$

4 0

2 years ago

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What is equivalent to 5/40

Mnenie [13.5K]

Is it a multiple choice question? I mean, 10/80?..

3 0

4 years ago

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Rewa Delta Union Rugby CEO has become concerned about the slow pace of the rugby games played in the current union rugby, fearin

ozzi

Answer:

We conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.

Step-by-step explanation:

We are given that Before the meeting, the mean duration of the 15-sided rugby game time was 3 hours, 5 minutes, that is, 185 minutes.

A random sample of 36 of the 15-sided rugby games after the meeting showed a mean of 179 minutes with a standard deviation of 12 minutes.

Let $\mu$ = mean duration of 15-sided union rugby games after the meeting.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 185 minutes {means that the mean duration of 15-sided union rugby games has increased or remained same after the meeting}

Alternate Hypothesis, $H_A$ : $\mu$ < 185 minutes {means that the mean duration of 15-sided union rugby games has decreased after the meeting}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean duration of 15-sided union rugby games = 179 min

s = sample standard deviation = 12 minutes

n = sample of 15-sided rugby games = 36

So, the test statistics = $\frac{179-185}{\frac{12}{\sqrt{36} } }$ ~ $t_3_5$

= -3

The value of t test statistics is -3.

Now, at 1% significance level the t table gives critical value of -2.437 at 35 degree of freedom for left-tailed test.

Since our test statistic is less than the critical value of t as -3 < -2.437, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.

6 0

4 years ago