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soldi70 [24.7K]
2 years ago
5

45 %of population of a town is female . what's the population of male if total population is 54320

Mathematics
2 answers:
Harman [31]2 years ago
5 0

Answer:29876

Step-by-step explanation:

Entire population is 100%

Male is 100%-45%

55%x54320=29876

Lubov Fominskaja [6]2 years ago
3 0

Answer:

29,876 males

Step-by-step explanation:

If 45% of 54320 population is female. The population of male is calculated below.

The total population is 100% and female is 45%. Thus: 100% - 45% = 55%.

Percentage of male population is 55%. Hence, 55% of 54320 is calculated thus:

\frac{55}{100}  \times 54320

= 0.55 \times 54320

= 29876

Therefore, population of male if total population is 54320 is 29,876.

Check,

Female population percentage + male population percentage:

45% + 55% = 100%

24,444 + 29,876 = 54,320

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75*x=90 \\ x= \frac{90}{75}  \\  \\ \boxed{x=1.2}

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4 0
2 years ago
The owners of Prim's Pizza are concerned that many of their customers are starting to purchase pizza from The Pizza Palace becau
AleksandrR [38]

Answer:

(240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953

(240-210) +1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=32.047

And the confidence interval for the difference is between:

27.953 \leq \mu_1 -\mu_2 \leq 32.047

Step-by-step explanation:

We have the following info given:

\bar X_1 = 240 sample mean for medium Pizzas from Prim's

\bar X_2 = 210 sample mean for medium Pizzas from Pizza Place

s_1 =8.6 sample deviation for Prim's

s_2 =5.7 sample deviation for Pizza Palca

n_1 =n_2 = 100  sample size selected for each case

The confidence interval for the difference of means is given by:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

And for the 95% confidence we need a significance level of \alpha=1-0.95=0.05 and \alpha/2 =0.025, the degrees of freedom are given by:

df= n_1 +n_2 -2= 100+100-2 =98

And the critical value would be

t_{\alpha/2}= 1.984

And replacing we got:

(240-210) -1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=27.953

(240-210) +1.984 \sqrt{\frac{8.6^2}{100}+\frac{5.7^2}{100}}=32.047

And the confidence interval for the difference is between:

27.953 \leq \mu_1 -\mu_2 \leq 32.047

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Answer:

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