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2 years ago

Find the area. 13mm 15mm 26mm 32mm ( PLEASE EXPLAIN or I give 2 stars)

Mathematics

2 answers:

Ber [7]2 years ago

8 0

You would just do 13+15+26+32 which would equal 86mm

vova2212 [387]2 years ago

5 0

623 !!!!!!!!!!!!!!!!!!!!!!

You might be interested in

HELP PLEASE.........

sleet_krkn [62]

Answer: $\bold{\dfrac{41}{80}}$

Step-by-step explanation:

$\dfrac{puppies}{total}+\dfrac{kittens}{total}+\dfrac{unicorns}{total}\\\\\\=\dfrac{15}{80}+\dfrac{16}{80}+\dfrac{10}{80}\\\\\\=\large\boxed{\dfrac{41}{80}}$

5 0

3 years ago

Clase, ¿cuánto es 2.000 − 502?

coldgirl [10]

Answer:

1498 or Mil cuatrocientos noventa y ocho (Espanol)

Step-by-step explanation:

2,000 - 502 = 2,000 - 500 - 2.

2,000 - 500 = 1500

1500 - 2 = 1498

6 0

3 years ago

Help!!! Brainliest to 1 or 2 answer!!

lyudmila [28]

Answer:

35-17=3y

Step-by-step explanation:

3y+17=35

-17 -17

3y=35-17

7 0

2 years ago

One True Love? A survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each pe

sertanlavr [38]

Answer:

99% confidence interval for the proportion of people who disagree with the statement is [0.667 , 0.713].

Step-by-step explanation:

We are given that a survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each person." has been conducted. The result is that 735 of the 2625 respondents agreed, 1812 disagreed, and 78 answered ‘‘don’t know."

Firstly, the pivotal quantity for 99% 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 people who disagree with the statement = $\frac{1812}{2625}$ = 0.69

n = sample of respondents = 2625

p = population proportion of people who disagree with statement

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.69-2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } }$ , $0.69+2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } }$ ]