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Strike441 [17]
3 years ago
12

60 boys at a summer camp were asked to complete a survey about their favorite outdoor activity. 18 boys chose biking and 29 boys

chose fishing. The remaining boys did not complete the survey. Which expression gives the percent of survey participants who chose fishing?
Mathematics
1 answer:
nadezda [96]3 years ago
8 0

━━━━━━━☆☆━━━━━━━

▹ Answer

\frac{29}{47} * 100

▹ Step-by-Step Explanation

29 + 18 = 47 → 29/47

Total boys → 100

29/47 * 100

Hope this helps!

CloutAnswers ❁

Brainliest is greatly appreciated!

━━━━━━━☆☆━━━━━━━

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Which number line shows the solution to the inequality −2(3x − 1) < 8?
Liula [17]

Answer:

x∈ (-1, ∞)

Step-by-step explanation:

<u>Solution of the inequality:</u>

  • −2(3x − 1) < 8            ⇒ Divide both sides by -2, this will change inequality    

                                              sign to opposite

  • 3x - 1 > -4                 ⇒ add 1 to both sides
  • 3x > -3                      ⇒ divide both sides by 3
  • x > -1                         ⇒ answer
  • x∈ (-1, ∞)                   ⇒ answer as set
4 0
3 years ago
Mya's lunchbox is shown. What is the volume of the lunchbox? Round to the nearest tenth if necessary.
NISA [10]
V= 720.8 cubed

Explanation- 4.25 x 2 = 8.5
H= 8.5
L= 10.6
W= 8
8.5 x 10.6 x 8 = 720.8
3 0
3 years ago
Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in
Leno4ka [110]

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the population variance is given as follows:

\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}

It is provided that:

<em>n</em> = 20

<em>s</em> = 3.9

Confidence level = 95%

⇒ <em>α</em> = 0.05

Compute the critical values of Chi-square:

\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}

\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45

Thus, the 95% confidence interval for the population variance is (8.80, 32.45).

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