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1 year ago

Does anyone know how to solve this?

Mathematics

1 answer:

Oksi-84 [34.3K]1 year ago

3 0

Answer: Okay so do 3x7.5 then 3x11.5 then 7.5x11.5 then add it up and you should get 143.25

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I don’t understand #6 question

romanna [79]

Answer:

B. (10, 2000)

Step-by-step explanation:

5000 - 300t = 1400ft - 1200f

=> t = 10

10 => (5000 - 300t) = 2000

So the answer is B. (10, 2000)

HOPE THIS HELPS AND HAVE A NICE DAY <3

6 0

1 year ago

Can you write this equation? #16

Oksana_A [137]

Equation: 7(3x-2)=133
Distribute: 21x-14=133
Add like terms: 21x=147
Divide by 21: x=7

4 0

3 years ago

Read 2 more answers

PLEASE HELP

Rudik [331]

$y=75\ \ \ \ x=5\\\\
y=?\ \ \ \ \ x=3\\\\
proportion:\\
\frac{75}{y}=\frac{5}{3}\ \ \ | cross\ multiplication\\\\
75*3=5y\\\\
225=5y\ \ \ | divide\ by\ 5\\\\
y=45
$

5 0

3 years ago

Plz help ill give u brainlist

PSYCHO15rus [73]

Hello!

It tells you what to do in the question

First round all the numbers to the nearest whole number

2.20 = 2
2.34 = 2
2.7 = 3

Then it says to multiply all the numbers

2 * 2 * 3 = 12

Then it says to divide by 6

12 / 6 = 2

The answer is 2 cubic yards

Hope this helps!

5 0

2 years ago

Read 2 more answers

Construct a 99% confidence interval for the population mean, mu. Assume the population has a normal distribution. A group of 1

Zarrin [17]

Answer:

99% confidence interval for the population mean is [19.891 , 24.909].

Step-by-step explanation:

We are given that a group of 19 randomly selected students has a mean age of 22.4 years with a standard deviation of 3.8 years.

Assuming the population has a normal distribution.

Firstly, the pivotal quantity for 99% 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 age of selected students = 22.4 years

s = sample standard deviation = 3.8 years

n = sample of students = 19

$\mu$ = population mean

Here for constructing 99% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.878 < $t_1_8$ < 2.878) = 0.99 {As the critical value of t at 18 degree of

freedom are -2.878 & 2.878 with P = 0.5%}

P(-2.878 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 2.878) = 0.99

P( $-2.878 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $2.878 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X -2.878 \times {\frac{s}{\sqrt{n} }$ , $\bar X +2.878 \times {\frac{s}{\sqrt{n} }$ ]

= [ $22.4 -2.878 \times {\frac{3.8}{\sqrt{19} }$ , $22.4 +2.878 \times {\frac{3.8}{\sqrt{19} }$ ]

= [19.891 , 24.909]

Therefore, 99% confidence interval for the population mean is [19.891 , 24.909].

6 0

3 years ago