All categories

2 years ago

0.376×800=(0.376×_)×100 =×100 =____

Mathematics

1 answer:

sergey [27]2 years ago

3 0

0.376*800

=(0.376*8)*100

=3.008*100

=300.8

You might be interested in

2. Write a function of the geometric sequence with a starting term of 10 and a common ratio of 4. Find the fourth term.

scoray [572]

Answer:last one pretty sure

Step-by-step explanation: sure

8 0

2 years ago

Help....... pls and thank u

grandymaker [24]

Answer:

9.42 cm

Step-by-step explanation:

Formula 2*3.14*r

2*3.14*radius

Find the raidus- 1.5

2*3.14*1.5 = 9.42 cm

5 0

3 years ago

If the exchange rate between the American dollar and the Japanese yen is such that $4.00= 442 yen, how many yen could be exchang

Anna [14]

$6(442Y/$4)=663Y

So $4 is equivalent to 663 Yen

5 0

3 years ago

Read 2 more answers

Is y= 2x cubed +6 linear or non-linear

zhuklara [117]

Non linear because nothing cubed or squared can be linear.
Hope this helps

6 0

2 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

0.376×800=(0.376×_____)×100 =______×100 =______

0.376×800=(0.376×_)×100 =×100 =____