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3 years ago

Write as a single power of 9 9^6 × 9^3 ÷ 9^4

Mathematics

2 answers:

Svetradugi [14.3K]3 years ago

6 0

Answer:

9^5

Step-by-step explanation:

9^6 × 9^3 9^9

_________ = _________ = 9^5

9^4 9^4

Have a good day and keep learning

bazaltina [42]3 years ago

3 0

Answer:

9^5is the answer.

Step-by-step explanation:

(i)9^6×9^3=9^9 when we multiply we will add the power.

(ii) 9^9÷9^4= 9^5 when we divide we will subtract the power.

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|90-(-20)|
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110 degrees celsius

3 0

3 years ago

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23.82600 round to three significant figures

fomenos

Answer:

23.8 i think

Step-by-step explanation:

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3 years ago

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Jenny and Natalie are selling cheesecakes for a school fundraiser. Customers can buy chocolate cakes and vanilla cakes. Jenny so

IrinaVladis [17]

The cost of 1 chocolate cake is $ 6 and cost of 1 vanilla cake is $ 7

Solution:

Let "c" be the cost of 1 chocolate cake

Let "v" be the cost of 1 vanilla cake

Jenny sold 14 chocolate cakes and 5 vanilla cakes for 119 dollars

Therefore, we can frame a equation as:

14 x cost of 1 chocolate cake + 5 x cost of 1 vanilla cake = 119

$14 \times c + 5 \times v=119$

14c + 5v = 119 ------- eqn 1

Natalie sold 10 chocolate cakes and 10 vanilla cakes for 130 dollars

Therefore, we can frame a equation as:

10 x cost of 1 chocolate cake + 10 x cost of 1 vanilla cake = 130

$10 \times c + 10 \times v = 130$

10c + 10v = 130 -------- eqn 2

Let us solve eqn 1 and eqn 2

Multiply eqn 1 by 2

28c + 10v = 238 ------ eqn 3

Subtract eqn 2 from eqn 3

28c + 10v = 238

10c + 10v = 130

( - ) --------------------------

18c = 108

c = 6

Substitute c = 6 in eqn 1

14(6) + 5v = 119

84 + 5v = 119

5v = 119 - 84

5v = 35

v = 7

Thus cost of 1 chocolate cake is $ 6 and cost of 1 vanilla cake is $ 7

8 0

3 years ago

Hurry i'll give brainiest if u help me

galben [10]

X^2 = b^2-c^2
x^2= sqrt117^2-6^2
x^2=sqrt117×sqrt117-36
x^2=117-36
x^2=81
x=sqrt81
x=9cm

6 0

3 years ago

A new security system needs to be evaluated in the airport. The probability of a person being a security hazard is 4%. At the ch

ivolga24 [154]

Answer:

(a) 0.9412

(b) 0.9996 ≈ 1

Step-by-step explanation:

Denote the events a follows:

$P$ = a person passes the security system

$H$ = a person is a security hazard

Given:

$P (H) = 0.04,\ P(P^{c}|H^{c})=0.02\ and\ P(P|H)=0.01$

Then,

$P(H^{c})=1-P(H)=1-0.04=0.96\\P(P|H^{c})=1-P(P|H)=1-0.02=0.98\\$

(a)

Compute the probability that a person passes the security system using the total probability rule as follows:

The total probability rule states that: $P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})$

The value of P (P) is:

$P(P)=P(P|H)P(H)+P(P|H^{c})P(H^{c})\\=(0.01\times0.04)+(0.98\times0.96)\\=0.9412$

Thus, the probability that a person passes the security system is 0.9412.

(b)

Compute the probability that a person who passes through the system is without any security problems as follows:

$P(H^{c}|P)=\frac{P(P|H^{c})P(H^{c})}{P(P)} \\=\frac{0.98\times0.96}{0.9412} \\=0.9996\\\approx1$

Thus, the probability that a person who passes through the system is without any security problems is approximately 1.

7 0

3 years ago