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3 years ago

7

Sally read 13 books during this month.her brother read 7 books.one fourth of all books they read were non fiction. how many non

fiction books did they read?

1 answer:

vladimir1956 [14]3 years ago

3 0

Thirteen plus seven is twenty. One fourth of twenty is five.
The answer is five.

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Which statements are true about the place values in the number 88.888? Choose all answers that are correct. A. The 8 in the hund

Vadim26 [7]

<span>Answers:
______________________________________________________

</span><span>[B]: The 8 in the tenths place is 10 times the value of the 8 in the hundredths place.
</span>______________________________________________________
<span>[C]: The 8 in the hundredths place is 10 times the value of the 8 in the tenths place.
______________________________________________________</span>

6 0

3 years ago

Read 2 more answers

If the relative frequency of a train being late is 0.15 how often could you expect the train to be late in 100 days?

neonofarm [45]

(100)= 0.25*100=15 times.

<span>SO, the train will be late 15 times in 100 days.

Hope i helped:P </span>

7 0

3 years ago

Determine whether the forces in the pair are pulling at right angles to each other.

snow_tiger [21]

Given:
right triangle
a = 2.9
c = 4.2

Pythagorean Theorem: a² + b² = c²

missing value of b.

b² = c² - a²
b² = (4.2)² - (2.9)²
b² = 17.64 - 8.41
b² = 9.23
b = √9.23
b = 3.03 or 3.0

The value of b is 3.0

4 0

3 years ago

What are the factors of the expression:<br><br> (y - 2)(x + 3)

timofeeve [1]

Answer:

(y - 2) and (x + 3)

Step-by-step explanation:

(y - 2)(x + 3) is already in factored form; the factors are (y - 2) and (x + 3).

3 0

3 years ago

An Xത chart with three-sigma limits has parameters as follows: UCL = 104; Center Line = 100; LCL = 96; n = 5 Suppose the process

ruslelena [56]

Answer:

The probability that the control chart would exhibit lack of control by at least the third point plotted is P=0.948.

Step-by-step explanation:

We consider "lack of control by at least the third point plotted" if at least one of the three first points is over the UCL or under the LCL.

The probability of one point of being over UCL=104 is:

$z=(X-\mu)/\sigma=(104-98)/8=6/8=0.75\\\\P(X>104)=P(z>0.75)=0.227$

The probability of one point of being under LCL=96 is:

$z=(X-\mu)/\sigma=(96-98)/8=-2/8=-0.25\\\\P(X$

Then, the probability of exhibit lack of control is:

$P=P(X>104)+P(X$

The probability of having at least one point out of control in the first three points is:

$P(x\geq1)=1-P(0)\\\\P(x\geq1)=1- 0.052=0.948\\\\\\P(x=0) = \binom{11}{0} p^{0}q^{3}= 1*1*0.0515=0.0515\\\\$

The probability that the control chart would exhibit lack of control by at least the third point plotted is P=0.948.

7 0

3 years ago