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3 years ago

Maths Assignment 4-a^2

Mathematics

2 answers:

Ugo [173]3 years ago

7 0

Answer:

Step-by-step explanation:

Use the identity: a² - b² = (a+ b)(a - b)

4 - a² = 2² - a² {Here a = 2 and b = a}

= (2 +a) (2 - a)

galben [10]3 years ago

5 0

Answer:

(2-a)^2

(2+a) (2-a)...

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Im struggling a bit can someone help me :)

Rasek [7]

Answer:

Step-by-step explanation:

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3 years ago

What’s the name of these mathematical symbols:

Alex

Answer:

he first is a numerical sentence and the second is an equation

Step-by-step explanation:

equation because there's "="

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3 years ago

In 1998, as an advertising campaign, the Nabisco Company announced a "1000 Chips Challenge," claiming that every 18-ounce bag of

Phoenix [80]

Answer:

A 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies is [1187.96, 1288.44].

Step-by-step explanation:

We are given that statistics students at the Air Force Academy (no kidding) purchased some randomly selected bags of cookies and counted the chocolate chips.

Some of their data are given below; 1219, 1214, 1087, 1200, 1419, 1121, 1325, 1345, 1244, 1258, 1356, 1132, 1191, 1270, 1295, 1135.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean number of chocolate chips = $\frac{\sum X}{n}$ = 1238.2

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 94.3

n = sample of car drivers = 16

$\mu$ = population mean number of chips in a bag

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.131 < $t_1_5$ < 2.131) = 0.95 {As the critical value of t at 15 degrees of

freedom are -2.131 & 2.131 with P = 2.5%}

P(-2.131 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.131) = 0.95

P( $-2.131 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.131 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.131 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $1238.2-2.131 \times {\frac{94.3}{\sqrt{16} } }$ , $1238.2+2.131 \times {\frac{94.3}{\sqrt{16} } }$ ]

= [1187.96, 1288.44]

Therefore, a 95% confidence interval for the mean number of chips in a bag of Chips Ahoy Cookies is [1187.96, 1288.44].

7 0

3 years ago

Ty has 2/6 of candy in tom has 3/8 of candy how much candy do they have in all

Anastasy [175]

To find out how much candy in all, add the two of the fractions.
2/6+3/8
2/6=1/3
1/3+3/8
Find the LCM of 3 and 8 and multiply the numerator and denominator by the same number.
LCM: 24
8/24+9/24= 17/24 altogether
Hope this helps.

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4 years ago

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A recipe calls for 2/3 cup of butter. Sticks of butter are divided into 1/4 cup sections. How many sections of butter will the r

Phoenix [80]

Step-by-step explanation:

They answer is your mom

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3 years ago

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