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4 years ago

Which statement is FALSE?

Mathematics

1 answer:

ad-work [718]4 years ago

7 0

Answer:

Step-by-step explanation:

This is only true if it is an isosceles trapezium so B is false.

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Write each number in word form and short form 9) 543,698 10) 45,987,159.9

Gemiola [76]

Answer:

9) five hundred and forty three thousand, six hundred and ninety-eight

500,000 + 40,00+ 3,000+ 600+ 90+ 8 = 543,698

10)forty-five million nine hundred eighty-seven thousand one hundred fifty-nine and nine tenths

40,000,000 + 5,000,000 + 900,000 + 80,000 + 7,000 + 100 + 50 + 9 + 0.9 = 45,987,159.9

8 0

3 years ago

HELP PLEASE ANYONE OR SOMEONE HELP. It’s due in 10 minutes.

ddd [48]

Sin V = opp/hyp
Sin 60 = opp/18
opp = sin 60 x 18
opp = 15.6
opp = 9 root 3

Answer:
D

6 0

3 years ago

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

Vladimir79 [104]

Answer: B

Step-by-step explanation:

Look at graph

4 0

3 years ago

A scientist determines that the number of bacteria in a culture is modeled by the formula A(t)=4,000e0.02t, where t is the numbe

SpyIntel [72]

Using the exponential model given, the number of bacteria in the culture after 9 minutes after the observation began will be 4789

Given the exponential model :

$A(t) = 4000e^{0.02t}$
$A(t) = Number\:bacteria\:in\:the\:culture\:after\:t\:minutes$
t = number of minutes after observation begins

The number of bacteria in the culture 9 minutes after the observation began can be calculated thus :

$t = 9 \: minutes$

Substitute 9 for t in the exponential equation :

$A(t) = 4000e^{0.02\times9}$

$A(t) = 4000e^{0.18}$

$A(t) = 4000 \times 1.1972173$

$A(t) = 4788.87$

Therefore, the number of bacteria in the culture after 9 minutes will be about 4789

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8 0

2 years ago

Final exam scores are normally distributed with a mean of 74 and a standard deviation of 6. Approximately, what percentage of fi

notka56 [123]

Answer:

81.86%

Step-by-step explanation:

We have been given that final exam scores are normally distributed with a mean of 74 and a standard deviation of 6.

First of all we will find z-score using z-score formula.

$z=\frac{x-\mu}{\sigma}$

$z=\frac{68-74}{6}$

$z=\frac{-6}{6}=-1$

Now let us find z-score for 86.

$z=\frac{86-74}{6}$

$z=\frac{12}{6}=2$

Now we will find P(-1<Z) which is probability that a random score would be greater than 68. We will find P(2>Z) which is probability that a random score would be less than 86.

Using normal distribution table we will get,

$P(-1$

$P(2>Z)=.97725$

We will use formula $P(a$ to find the probability to find that a normal variable lies between two values.

Upon substituting our given values in above formula we will get,

$P(-1$

Upon converting 0.81859 to percentage we will get

$0.81859*100=81.859\approx 81.86$

Therefore, 81.86% of final exam score will be between 68 and 86.

3 0

3 years ago