All categories

3 years ago

What’s 0.604 in word from

Mathematics

2 answers:

Leni [432]3 years ago

7 0

six hundred four thousandths

~ or, simpler ~

zero point six hundred four

~ or, even simpler ~

zero point six zero four

Elenna [48]3 years ago

5 0

Answer:

six hundred four thousandths

You might be interested in

What is the answer to 9x(6+x)

Anna11 [10]

$54x + x \\ \\ (constributive \: property)$

5 0

3 years ago

Could someone give me an answer and explain how you got it ?

tankabanditka [31]

Answer:

A (9, 3)

Step-by-step explanation:

First the point is rotated 90° counterclockwise about the origin. To do that transformation: (x, y) → (-y, x).

So S(-3, -5) becomes S'(5, -3).

Next, the point is translated +4 units in the x direction and +6 units in the y direction.

So S'(5, -3) becomes S"(9, 3).

3 0

3 years ago

Let P be a point between points S and T on 2004-01-01-02-00_files/i0120000.jpg. If ST = 21, SP = 3b – 11, and PT = b + 4, solve

NNADVOKAT [17]

The best and most correct answer among the choices provided by your question is the third choice or letter C.

If ST = 21, SP = 3b – 11, and PT = b + 4, the value of b would be 14.


I hope my answer has come to your help. God bless and have a nice day ahead!

7 0

3 years ago

Read 2 more answers

A statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after

lana [24]

Answer:

95% confidence interval estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

(a) Lower Limit = 0.486

(b) Upper Limit = 0.624

Step-by-step explanation:

We are given that a statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after receiving their bachelor's.

She took a random sample of 200 graduates from the class of 1979 and determined their occupations in 1989. She found that 111 persons were still employed primarily as engineers.

Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of persons who were still employed primarily as engineers = $\frac{111}{200}$ = 0.555

n = sample of graduates = 200

p = population proportion of engineers

Here for constructing 95% confidence interval we have used One-sample z proportion test statistics.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.555-1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } }$ , $0.555+1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } }$ ]