All categories

3 years ago

Which expression has the greatest value? A.-5+(-8) B.-6+-4 .C.-4+6 D.-4+-6

Mathematics

1 answer:

Snowcat [4.5K]3 years ago

4 0

Answer: C
Explanation:
A=-13
B=-10
C=2
D=-10
Therefore, C is the greatest.
I hope this helps! :)

You might be interested in

Can anyone help me with this question??

zepelin [54]

Answer:

The asnwer is 83.6 :)

Step-by-step explanation:

7.6x11=83.6

4 0

3 years ago

Read 2 more answers

Which solid has six faces, four lateral faces, two bases, eight vertices, and 12 edges? square pyramid triangular prism rectangu

marusya05 [52]

Rectangular Prism
*********************************************

4 0

3 years ago

Read 2 more answers

Brandon gets a settlement from a car accident. The other driver will be paying him $2000 each month for the next 5 years to cove

Natalija [7]

Answer:

That would be $120,000 at the end of the 5 years.

Step-by-step explanation:

$2,000 x 12 months = $24,000

$24,000 x 5 years = $120,000 total

4 0

3 years ago

Teen obesity:

Law Incorporation [45]

Answer:

95% confidence interval for the percentage of all U.S. public high school students who are obese is [0.110 , 0.190].

Step-by-step explanation:

We are given that 15% of a random sample of 300 U.S. public high school students were obese.

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 % of U.S. public high school students who were obese = 15%

n = sample of U.S. public high school students = 300

p = population percentage of all U.S. public high school students

Here for constructing 95% confidence interval we have used One-sample z proportion 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 = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.15-1.96 \times {\sqrt{\frac{0.15(1-0.15)}{300} } }$ , $0.15+1.96 \times {\sqrt{\frac{0.15(1-0.15)}{300} } }$ ]

= [0.110 , 0.190]

Therefore, 95% confidence interval for the percentage of all U.S. public high school students who are obese is [0.110 , 0.190].

6 0

3 years ago

What is the absolute value of negative 75? ∣−75∣ =?

zzz [600]

the absolute value is positive 75.

5 0

2 years ago

Read 2 more answers