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3 years ago

14

You measure 27 watermelons' weights, and find they have a mean weight of 71 ounces. Assume the population standard deviation is

2.8 ounces. Based on this, construct a 95% confidence interval for the true population mean watermelon weight.

1 answer:

velikii [3]3 years ago

5 0

Answer:

(69.94, 72.06)

Step-by-step explanation:

Confidence Interval formula = Mean ± z × Standard deviation/√n

Mean = 71 ounces

Standard deviation = 2.8 ounces

z score = 95% confidence interval = 1.96

Confidence Interval = 71 ± 1.96 × 2.8/√27

= 71 ± 1.96 × 0.5388602512

= 71 ± 1.0561660924

71 - 1.0561660924

= 69.943833908 ounces

Approximately = 69.94 ounces

71 + 1.0561660924

= 72.0561660924 ounces

Approximately = 72.06 ounces

95% Confidence Interval = (69.94, 72.06).

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Pls help i will give you a 5 star rating

docker41 [41]

Question # 1

Given the expression

$6^2\div \:3+\left(5+3\cdot \:2\right)-2^3$

Follow the PEMDAS order of operations

$\mathrm{Calculate\:within\:parentheses}\:\left(5+3\cdot \:2\right)\::\quad 11$

$=6^2\div \:3+11-2^3$

$\mathrm{Calculate\:exponents}\:6^2\::\quad 36$

$=36\div \:3+11-2^3$

$\mathrm{Calculate\:exponents}\:2^3\::\quad 8$

$=36\div \:3+11-8$

$\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:36\div \:3\::\quad 12$

$=12+11-8$

$\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:12+11-8\::\quad 15$

$=15$

Therefore,

$6^2\div \:3+\left(5+3\cdot \:2\right)-2^3=15$

Question # 2

Given the expression

$\frac{4+3^2-15\div 5}{\left(2^4-5\cdot \:3\right)^2}$

as

$4+3^2-\frac{15}{5}$

$=4+9-\frac{15}{5}$ ∵ $3^2=9$

$\mathrm{Add\:the\:numbers:}\:4+9=13$

$=-\frac{15}{5}+13$

and

$\left(2^4-5\cdot \:3\right)^2$

$=\left(16-5\cdot \:3\right)^2$ ∵ $2^4=16$

$\mathrm{Multiply\:the\:numbers:}\:5\cdot \:3=15$

$=\left(16-15\right)^2$

$\mathrm{Subtract\:the\:numbers:}\:16-15=1$

$=1^2$

$\mathrm{Apply\:rule}\:1^a=1$

$=1$

Thus the equation $\frac{4+3^2-15\div 5}{\left(2^4-5\cdot \:3\right)^2}$ becomes

$=\frac{-\frac{15}{5}+13}{1}$

$\mathrm{Divide\:the\:numbers:}\:\frac{15}{5}=3$

$=\frac{-3+13}{1}$

$\mathrm{Apply\:rule}\:\frac{a}{1}=a$

$=-3+13$

$\mathrm{Add/Subtract\:the\:numbers:}\:-3+13=10$

$=10$

Therefore,

$\frac{4+3^2-\frac{15}{5}}{\left(2^4-5\cdot \:3\right)^2}=10$

Question # 3

Given the expression

$ab-c^2+2b$

Putting a = 2, b = 4, and c = 1 in the expression

$=\left(2\right)\left(4\right)-\left(1\right)^2+2\left(4\right)$

Follow the PEMDAS order of operations

$\mathrm{Calculate\:exponents}\:\left(1\right)^2\::\quad 1$

$=\left(2\right)\left(4\right)-1+2\left(4\right)$

$\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:\left(2\right)\left(4\right)\::\quad 8$

$=8-1+2\left(4\right)$

$\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:2\left(4\right)\::\quad 8$

$=8-1+8$

$\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:8-1+8\::\quad 15$

$=15$

Therefore,

$ab-c^2+2b=\left(2\right)\left(4\right)-\left(1\right)^2+2\left(4\right)=15$

Question # 4

Given the expression

$4d^3+2e\div \:f+de$

Putting d = 2, e = 3, and f = 6 in the expression

$=4\left(2\right)^3+2\left(3\right)\div \:6+\left(2\right)\left(3\right)$

Follow the PEMDAS order of operations

$\mathrm{Calculate\:exponents}\:\left(2\right)^3\::\quad 8$

$=4\cdot \:8+2\left(3\right)\div \:6+\left(2\right)\left(3\right)$

$\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:4\cdot \:8\::\quad 32$

$=32+2\left(3\right)\div \:6+\left(2\right)\left(3\right)$

$\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:2\left(3\right)\div \:6\::\quad 1$

$=32+1+\left(2\right)\left(3\right)$

$\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:\left(2\right)\left(3\right)\::\quad 6$

$=32+1+6$

$\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:32+1+6\::\quad 39$

$=39$

Therefore,

$4d^3+2e\div \:\:f+de=4\left(2\right)^3+2\left(3\right)\div \:6+\left(2\right)\left(3\right)=39$

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Step-by-step explanation:

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