All categories

1 year ago

PLSSS HELP IF YOU TURLY KNOW THISS

Mathematics

1 answer:

balandron [24]1 year ago

5 0

Step-by-step explanation:

we have 11k + 4k.

this is like saying "I have 11 apples and then get 4 more apples - how many apples do I have ?"

in other words, how many "k" do I have ?

11 + 4 = 15.

we have 15k.

and then there is -3 + 8 = 5

so, the answer is

15k + 5

You might be interested in

6q − 18 > 5q − 23 solve

frez [133]

Answer:

6q-2=66.1

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

HELP URGENT MATH PROBLEM I WILL GIVE BRAINLIEST AND 5 STAR RATING

zhannawk [14.2K]

Day = 8 4/5 ÷ 4
day = 44/5 ÷ 4
day = 44/5 × 1/4
day = 44/20
day = 11/5
day = 2 1/5

6 0

3 years ago

)which of the following correctly uses absolute value to show the distance between −70 and 15? |−70 − 15| = |−85| = −85 units |−

Crazy boy [7]

|−70 − 15| = |−85| = 85 units

3 0

3 years ago

Read 2 more answers

Find the volume of the pyramid. v=

sladkih [1.3K]

Answer:

240 mi^3

Step-by-step explanation:

Base area = length x width = 10 x 8 = 80 mi^2

V = (base area x height)/3 = (80 x 9)/3 = 240 mi^3

5 0

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