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4 years ago

HELP MEEEE!! NEED ANSWERE ASAP!!!!

Mathematics

2 answers:

Stells [14]4 years ago

3 0

The first one is -8+2 = -6

And the second one is 2 x 2 = 4

Zinaida [17]4 years ago

3 0

Answer:

Step-by-step explanation:

Total is -6

-24 x 4

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Substitute -3x-8y=4 -2x+7y=15

Nikolay [14]

-2x=15-7y
x=-7.5+3.5y

-3x-8y=4
-3(7.5+3.5y)-8y=4
-22.5-10.5y-8y=4
-22.5-18.5y=4
-18.5y=4+22.5
-18.5y=26.5
y=-53/35

the fraction might not be in lowest terms.

3 0

3 years ago

I need help :) Find the GCF: 10x^5−16x^4+4x^2

andreev551 [17]

The greatest common factor is 2x^2.
the largest number that divides evenly into 10x^5−16x^4+4x^2 is two.
the highest degree is x^2 so the answer would be 2x^2

5 0

3 years ago

A rectangle has a

adelina 88 [10]

Answer:

3x + 6y

Step-by-step explanation:

so the perimeter is 12x + 8y and a side is 4x-2y so that means that side and the side across is 8x-4y so you put it all to one side to get 6x +12y then divide that by 2 to find 1 side so the side is 3x + 6y

8 0

2 years ago

Lauren read a total of 44 books over 11 months. After belonging to the book club for 21 months, how many books will Lauren have

GaryK [48]

Answer:

Step-by-step explanation:

Lauren has read 4 books per month. So the answer would be 84.

6 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

**HELP MEEEE!! NEED ANSWERE ASAP!!!!**

HELP MEEEE!! NEED ANSWERE ASAP!!!!