Which two numbers add up to _ and multiply to _?

1 answer:

3 0

1. multiply to -18 add to -17 = Multiplying -18 and 1 will give -18 as the result and then on adding -18 and + 1 the result comes to -17
2. multiply to 36 add to -13 = Multiplying -9 and -4 will give -36 as the result and then on adding -9 and -4 the result comes to -13
3. multiply to -24 add to -5= Multiplying -8 and +3 will give -24 as the result and then on adding -8 and +3 the result comes to -5
4. multiply to -18 add to 7= Multiplying +9 and -2 will give -18 as the result and then on adding +9 and -2 the result comes to 7
5. multiply to -36 add to 9= Multiplying +12 and -3 will give -36 as the result and then on adding +12 and -3 the result comes to 9
 6. multiply to 24 add to 10= Multiplying +6 and +4 will give 24 as the result and then on adding +6 and +4 the result comes to 10
 7. multiply to 18 add to -9= Multiplying -6 and -3 will give -18 as the result and then on adding -6 and -3 the result comes to -9
 8. multiply to -36 add to -16= Multiplying -18 and +2 will give -36 as the result and then on adding -18 and +2 the result comes to -16

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Answer:

95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

Step-by-step explanation:

We are given that during the 7th examination of the Offspring cohort in the Framing ham Heart Study, there were 1219 participants being treated for hypertension and 2,313 who were not on treatment.

The sample proportion is : $\hat p$ = x/n = 1219/3532 = 0.345

Firstly, the pivotal quantity for 95% confidence interval for the proportion of the population 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 = 0.345

n = sample of participants = 3532

p = population proportion

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

= [ $0.345-1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ , $0.345+1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ ]