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2 years ago

8-1-(18-2)/18 /= divide use pemdas

Mathematics

1 answer:

Blizzard [7]2 years ago

7 0

The correct answer is 0.5, I hoped this helped! :)

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Hey there! Can you please help me with 6th grade :-)

Brilliant_brown [7]

Oh i miss 6th grade im in 7th now lol and yes but i thank number 2 is 3 cnts

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2 years ago

What is number 5???!!

vagabundo [1.1K]

The value of m∠M is 112°

Step-by-step explanation:

In the parallelogram NPMQ, opposite angles ∠P=∠M.

To get the value of the angles, equate the two expressions;

(6x-2)°=(4x+36)°

6x-2=4x+36 ------ collect like terms

6x-4x=36+2

2x=38

x=38/2 =19

So, ∠M= (6x-2)° = (6×19 - 2)° = 112°

Learn More

Angles in a parallelogram : brainly.com/question/11611093

Keyword : angles

#LearnwithBrainly

8 0

3 years ago

Bárbara come la tercera parte de una caja de bombones. Su papá luego compra 7 bombones más y los pone en la caja. ¿Cómo se repre

Maksim231197 [3]

Answer:

Bárbara come la tercera parte de una caja de bombones. Su papá luego compra 7 bombones más y los pone en la caja. ¿Cómo se representa la cantidad de bombones que hay ahora en la caja?

Step-by-step explanation:

3 0

2 years ago

Need help asap!! In the picture above!!

mylen [45]

The last one is the right answer b/c 7 plus 0.30 for each topping, -> t

4 0

3 years ago

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

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

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

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

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

= [19.891 , 24.909]

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

6 0

3 years ago