Answer:
35km
Step-by-step explanation:
1:500000
7:?
(cross multiply)
=3500000cm
1km=100000cm
3500000cm
<u>3</u><u>5</u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>c</u><u>m</u><u>×</u><u>1</u><u>k</u><u>m</u>
100000cm
=35km
Answer:
9x - 2x + 62 = 9x + 24
Number of correct answer each obtained = 19
Step-by-step explanation:
Given that:
Number of correct answers = x
POINTS for correct answers = 9
POINTS for incorrect answers = - 2
SCHOOL A:
Number of correct answers = x
Number of incorrect answers = x
Initial point = 62
Initial point + 9(number of correct answers) + - 2(number of incorrect answers)
62 + 9x - 2x
SCHOOL B :
Number of correct answers = x
Number of incorrect answers = 0
Initial point = 24
Initial point + 9(number of correct answers) + - 2(number of incorrect answers
24 + 9x
Since they end up tied :
School A = School B
62 + 9x - 2x = 24 + 9x
9x - 2x + 62 = 9x + 24
Number of correct answers gotten :
9x - 2x + 62 = 9x + 24
7x + 62 = 9x + 24
7x - 9x = 24 - 62
-2x = - 38
x = 19
Hence, Number of correct answers each obtained = 19
answer:
1.57 cm
Step-by-step explanation:
pi = 3.14
circumference of circle = pi x diameter
3.14 x 2 = 6.28
90/360 x 6.28 = 1.57
the arc length = 1.57 provided that 3.14 is used as pi.
Answer:$125
Step-by-step explanation:
50% of 250
50/100 x 250
1/2 x 250=(1x250)/2=250/2=125
Answer:
Step-by-step explanation:
Hello!
The variable of interest is:
X: number of daily text messages a high school girl sends.
This variable has a population standard deviation of 20 text messages.
A sample of 50 high school girls is taken.
The is no information about the variable distribution, but since the sample is large enough, n ≥ 30, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:
X[bar]≈N(μ;δ²/n)
This way you can use an approximation of the standard normal to calculate the asked probabilities of the sample mean of daily text messages of high school girls:
Z=(X[bar]-μ)/(δ/√n)≈ N(0;1)
a.
P(X[bar]<95) = P(Z<(95-100)/(20/√50))= P(Z<-1.77)= 0.03836
b.
P(95≤X[bar]≤105)= P(X[bar]≤105)-P(X[bar]≤95)
P(Z≤(105-100)/(20/√50))-P(Z≤(95-100)/(20/√50))= P(Z≤1.77)-P(Z≤-1.77)= 0.96164-0.03836= 0.92328
I hope you have a SUPER day!