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4 years ago

9

If m angle BGF=152 degrees, what is m angle AGF

2 answers:

PtichkaEL [24]3 years ago

8 0

Answer:

A, #1, uno, first one

Step-by-step explanation:

Allushta [10]4 years ago

4 0

The Answer is: 64 degrees

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Evaluate 5(14 - 2³) +8 ÷ 2.? Plz help me, will mark you as brainliest A. 26 B. 34 C. 44 D. 56

dangina [55]

Answer:

$\large \boxed{\sf B. \ 34}$

Step-by-step explanation:

$5(14 - 2^3 ) +8 \div 2$

$\sf Solve \ brackets.$

$2^3 =8$

$5(14 - 8 ) +8 \div 2$

$14-8=6$

$5(6 ) +8 \div 2$

$\sf Multiply \ and \ divide.$

$5(6)=30$

$8 \div 2 = 4$

$30 +4$

$\sf Add.$

$34$

3 0

3 years ago

Read 2 more answers

StephanieHad 7/8 pound of bird seed she can use 3/8 pound to fill a bird feeder how much birdseed to Stephanie have left

Anna11 [10]

7/8 - 3/8= 4/8 I hope this helps.

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4 years ago

ONLY ANSWER IF U KNOW!<br>Find the limit of the function algebraically.

blsea [12.9K]

Answer:

lim [(x(2) + 3) × 1/ x(4) ] = 3 × ( + ○○) = + ○○

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3 years ago

Samuel and his family water-ski on the Allegheny River. If their boat’s speedometer reads 16 mi/h and Sam stays on his skis for

storchak [24]

60÷18 = 0.3

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3 years ago

A home security system is designed to have a 99% reliability rate. Suppose that nine homes equipped with this system experience

bulgar [2K]

Answer:

a) $P(X \geq 1) = 1-P(X$

$P(X=0)=(9C0)(0.99)^0 (1-0.99)^{9-0}=1x10^{-18}$

And replacing we got:

$P(X \geq 1) =1 -1x10^{-18} \approx 1$

b) $P(X=7)=(9C7)(0.99)^7 (1-0.99)^{9-7}=0.003355$

$P(X=8)=(9C8)(0.99)^8 (1-0.99)^{9-8}=0.083047$

$P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517$

And adding we got:

$P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992$

c) $P(X \leq 8) =1 -P(X>8) = 1-P(X=9)$

$P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517$

And replacing we got:

$P(X \leq 8)= 1-0.913517=0.086483$

Step-by-step explanation:

Let X the random variable of interest "numebr of times that an alarm is triggered", on this case we now that:

$X \sim Binom(n=9, p=0.99)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want to find this probability:

$P(X \geq 1) = 1-P(X$

$P(X=0)=(9C0)(0.99)^0 (1-0.99)^{9-0}=1x10^{-18}$

And replacing we got:

$P(X \geq 1) =1 -1x10^{-18} \approx 1$

Part b

$P(X \geq 7)= P(X=7) +P(X=8)+ P(X=9)$

$P(X=7)=(9C7)(0.99)^7 (1-0.99)^{9-7}=0.003355$

$P(X=8)=(9C8)(0.99)^8 (1-0.99)^{9-8}=0.083047$

$P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517$

And adding we got:

$P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992$

Part c

$P(X \leq 8) =1 -P(X>8) = 1-P(X=9)$

$P(X=9)=(9C9)(0.99)^9 (1-0.99)^{9-9}=0.913517$

And replacing we got:

$P(X \leq 8)= 1-0.913517=0.086483$

4 0

3 years ago