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Vaselesa [24]
3 years ago
15

17 + x = 18 in 6th grade math what’s the answer?

Mathematics
2 answers:
Masteriza [31]3 years ago
6 0
Subtract 17 from both sides , x=1
morpeh [17]3 years ago
6 0
Hello!!! Hope you are having a good day!!! So for this type of equation just do 18-17=1 so x=1. This is because the equation they are just asking 17 plus what equals 18 and 17+1=18
Hope this helps have a good day!!
(I only need one more brainliest, so it will be appreciated if I get one!!!) :))))
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It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose you randomly survey
mote1985 [20]

Answer:

P(X\geq 8)=0.0043\\\\

It's more likely that  all of the residents surveyed will have adequate earthquake supplies since it has a probability of 98.02% which is very close to 100%.

Step-by-step explanation:

-This is a binomial probability problem with the function:

P(X=x)={n\choose x}p^x(1-p)^{n-x}

-Given p=0.3, n=11, the is calculated as:

P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 8)=P(X=8)+P(X=9)+P(X=10)+P(X=11)\\\\={11\choose 8}0.3^8(0.7)^3+{11\choose 9}0.3^9(0.7)^2+{11\choose 10}0.3^{10}(0.7)^1+{11\choose 11}0.3^{11}(0.7)^0\\\\=0.0037+0.0005+0.00005+0.000002\\\\=0.0043

Hence, the probability that at least 8 have adequate supplies 0.0043

#The probability that non has adequate supplies is calculated as;

P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X= 0)={11\choose 0}0.3^{0}(0.7)^{11}\\\\=0.0198

#The probability that all have adequate supplies is calculated as:

P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X= All)=1-{11\choose 0}0.3^{0}(0.7)^{11}\\\\=1-0.0198\\\\=0.9802

Hence, it's more likely that  all of the residents surveyed will have adequate earthquake supplies since P(All)>P(None)\ \ and that this probability is 0.9802 or 98.02% a figure close to 1

7 0
3 years ago
Someone please solve this for me!
mezya [45]

Answer:

  x = 149

Step-by-step explanation:

The transversal <em>t</em> crosses parallel lines <em>m</em> and <em>n</em> creating a number of angles. Both marked angles are above and to the right of the point of intersection, so they are <em>corresponding angles</em>.

Corresponding angles in this geometry have identical measures:

  x° = 149°

  x = 149

6 0
3 years ago
The firm is requested to send 3 employees who have positive indications of asbestos on to a medical center for further testing.
Rus_ich [418]

Answer:

0.0645 = 6.45% probability that exactly 10 employees will be tested in order to find 3 positives

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

40% of the employees have positive indications of asbestos in their lungs

This means that p = 0.4

a) Find the probability that exactly 10 employees will be tested in order to find 3 positives

2 within the first 9(P(X = 2) when n = 9), and the 10th, with 0.4 probability. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{9,2}.(0.4)^{2}.(0.6)^{7} = 0.1612

0.4*0.1612 = 0.0645

0.0645 = 6.45% probability that exactly 10 employees will be tested in order to find 3 positives

4 0
3 years ago
What is 66 tens + 24 tens
natita [175]

Answer:

900.

Step-by-step explanation:

66 tens are 660.

and 24 tens are 240.

so, It is 900!

3 0
3 years ago
(3.2 * 10000000) -3
andreev551 [17]

The exponents distribute over factors:

(ab)^c = a^c \times b^c

so, you have

(3.2 \times 10^6)^{-3} = 3.2^{-3} \times (10^6)^{-3}

Let's focus on each factor: applying the definition of negative exponents, you have

3.2^{-3} = \cfrac{1}{3.2^3} \approx 0.03

While for the power of 10, you can use the following rule

(a^b)^c = a^{bc} to write

(10^6)^{-3} = 10^{6\cdot (-3)} = 10^{-18}

So, the expression evaluates to

0.03\times 10^{-18} = 3\times 10^{-20}

3 0
4 years ago
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