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Tems11 [23]
2 years ago
15

598+20+987+78+90+87+80+60+70+90​

Mathematics
1 answer:
aleksklad [387]2 years ago
5 0

Answer:

2,160

Step-by-step explanation:

598 + 20 = 618

618 + 987 = 1,605

1,605 + 78 = 1,683

1,683 + 90 = 1,773

1,773 + 87 = 1,860

1,860 + 80 = 1,940

1,940 + 60 = 2,000

2000 + 70 = 2,070

2070 + 90 = 2,160

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The rate of change is 4.A linear model
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irakobra [83]
X = 42 - 19, so A - x = 23
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Read 2 more answers
Rewrite the rational exponent as a radical by extending the properties of integer exponents. 2 7/8 divided by 2 1/4
VladimirAG [237]
We know the rule of change: a b/c= (ac+b):c
So we have 2 7/8= (2x8+7):8= 23/8
and the second: 2 1/4 = (2x4+1):4= 9/4
We do the division: 23/8: 9/4= 23/8 x 4/9= (23x4):(8x9)= 23/18= 1  5/18
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3 years ago
Can someone please help me i dont understand this at all
goblinko [34]

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D.

40 x 6.4 = 256

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Step-by-step explanation:

5 0
3 years ago
A manufacturer of processing chips knows that 2\%2%2, percent of its chips are defective in some way. Suppose an inspector rando
kipiarov [429]

The data in the question seems a bit erroneous. I am writing the correct question below:

A manufacturer of processing chips knows that 2%, percent of its chips are defective in some way. Suppose an inspector randomly selects 4 chips for an inspection. Assuming the chips are independent, what is the probability that at least one of the selected chips is defective? Lets break this problem up into smaller pieces to understand the strategy behind solving it.

Answer:

The probability that at least one of the selected chips is defective is 0.0776.

Step-by-step explanation:

The question states that the probability of defective chips is 2% i.e. 0.02. Let p denote the probability of selecting a defective chip so, p = 0.02

An inspector selects 4 chips, which means n=4 and we need to compute the probability that at least one of the selected chips is defective. Let X be the number of defective chips selected. We need to compute P(X≥1) which means either 1, 2, 3 or 4 chips can be defective.

We will use the binomial distribution formula to solve this problem. The formula is:

<u>P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ</u>

where n = total no. of trials

          p = probability of success

          x = no. of successful trials

          q = probability of failure = 1-p

we have n=4, p=0.02 and q=1-0.02=0.98.

We need to compute P(X≥1) which is equal to:

P(X≥1) = P(X=1) + P(X=2) + P(X=3) + P(X=4)

A shorter method to do this is to use the total probability theorem:

P(X≥1) = 1 - P(X<1)

          = 1 - P(X=0)

          = 1 - ⁴C₀ (0.02)⁰(0.98)⁴⁻⁰

          = 1 - (0.98)⁴

          = 1 - 0.9224

P(X≥1) = 0.0776

4 0
3 years ago
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