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3 years ago

19 x 19^3 solve and answer

Mathematics

2 answers:

Mrrafil [7]3 years ago

8 0

Answer:

$\huge\boxed{Answer\hookleftarrow}$

$19 \times {19}^{3} \\ = 19 \times (19 \times 19 \times 19) \\ = 19 \times 19 \times 19 \times 19 \\ = 130321$

ʰᵒᵖᵉ ⁱᵗ ʰᵉˡᵖˢ

# ꧁❣ RainbowSalt2²2² ࿐

Viefleur [7K]3 years ago

6 0

Answer:

130321

Step-by-step explanation:

branliest????

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Please hurry find the measure of P

Inessa [10]

Answer:

m∠P = 65°

Step-by-step explanation:

An Isosceles triangle has two sides of equal length and two interior angles of equal measure.

Sum of interior angles of a triangle = 180°

If QO ≅ PQ and the enclosed m∠Q = 50° then m∠P ≅ m∠O

⇒ m∠Q + m∠P + m∠O = 180

⇒ 50 + m∠P + m∠O = 180

⇒ m∠P + m∠O = 130

As m∠P ≅ m∠O:

⇒ m∠P = 130 ÷ 2 = 65°

7 0

2 years ago

Read 2 more answers

During the 7th examination of the Offspring cohort in the Framingham Heart Study, there were 1219 participants being treated for

AlexFokin [52]

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

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

= [0.3293 , 0.3607]

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

6 0

3 years ago

Can you show me how to solve-8/2

Mrrafil [7]

-8/2

divide -8 by 2

-8 divided by 2 = -4

6 0

3 years ago

WHAT IS 9+8+9-2+5x5+5-25+5-4+2+-2+4-+2-5+4-2+5-4+2-5+4-1+5-4+-2+4-2+5-4+2-4 +-+

a_sh-v [17]

Answer:

Step-by-step explanation:

cause you had to doit the simple math problem

are you trying to prank on this problem

8 0

4 years ago

Find the number of ways of arranging the numbers

Doss [256]

First of all, note that all integers are either 0,1, or 2 modulo 3 (if you're not familiar with this terminology, it means that every integer is either a multiple of 3, or it is 1 or 2 away from a multiple of 3).

So, we can think of our numbers as

$\begin{array}{c|c}x&x\mod 3\\0&0\\1&1\\2&2\\3&0\\4&1\\5&2\\6&0\\7&1\\8&2\\9&0\end{array}$

In order to make sure that the sum of any three adjacent numbers is divisible by 3, we have to make sure that any group of 3 three adjacent numbers contains a 0, a 1 and a 2. This is possible only if we arrange our 9 numbers in 3 groups of 3 numbers containing 0,1 and 2 exactly once, repeating always the same pattern.

For example, we could arrange our numbers following the pattern

$0,1,2,0,1,2,0,1,2$

$2,0,1,2,0,1,2,0,1$

We have $3!=6$ possible patterns. Suppose for example that we choose the pattern

$0,1,2,0,1,2,0,1,2$

One possible way of following this pattern would be the arrangement

$3,1,2,6,4,5,9,7,8$

In fact, we substituted every '0' with a multiple of 3 (3, 6 or 9), every '1' with a number 1 away from a multiple of 3 (1, 4 or 7) and every '2' with a number 2 away from a multiple of 3 (2, 5 or 8).

This means that, once we fix a patter, we have 3 choices for the first 3 slots, 2 choices for the next 3 slots, and the final slot will be fixed. So, we have

$3\cdot 3\cdot 3\cdot 2 \cdot 2 \cdot 2 = 216$

possible ways of following a fixed pattern. Since the number of patterns was 6, we have

$216\cdot 6 = 1296$

possible arrangements.

7 0

3 years ago

Read 2 more answers