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3 years ago

Helppppp i need help ASAP

Mathematics

1 answer:

algol133 years ago

4 0

Answer:

a) 10.10 a.m. , (b) 10.20 a.m. which is 10 minutes after sprints start , (c) The students work at 10.50 a.m. , (d) For 60 minutes from 10 a.m. to 11 a.m. which equals to 1 hour

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The arena at the Berry Center holds 9,500 people. During one of the graduation ceremonies, 95% of the seats were occupied. How m

motikmotik

Answer:

9,025

Step-by-step explanation:

9,500 multiplied by 0.95

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3 years ago

The global minimum of the function f(x)

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3 years ago

What average speed does he need to reach in his second round in order to attain the overall average speed of 60 mph required for

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3 years ago

A study sampled 350 upperclassmen (Group 1) and 250 underclassmen (Group 2) at high schools around the city of Houston. The stud

iris [78.8K]

Answer:

$-0.048$

Do not reject $H_0:P_1-P_2=0$

Step-by-step explanation:

From the question we are told that

Sample size $n_1=350$

Sample size $n_2=250$

Sample proportion 1 $\hat P= \frac{25}{350} =>0.07$

Sample proportion 2 $\hat P= \frac{19}{250} =>0.076$

95% confidence interval

Generally for 95% confidence level

Level of significance

$\alpha = 1-0.95=>0.05$

$\alpha /2=\frac{0.05}{2} =>0.025$

Therefore

$Z_a_/_2=1.96$

Generally the equation for confidence interval between $P_1 - P_2$ is mathematically given as

$(\hat P_1-\hat P_2)\pm Z_a_/_2\sqrt{\frac{\hat P_1(1-\hat P_1)}{n_1}+\frac{\hat P_2(1-\hat P_2)}{n_2} }$

$(0.07-0.076)\pm 1.96\sqrt{\frac{0.07(1-0.07)}{350}+\frac{0.076(1-0.076)}{250} }$

$(0.07-0.076)\pm 1.96\sqrt{4.66896*10^-^4 }$

$(-0.006)\pm 0.042$

$(-0.006)- 0.042=>-0.048$

$(-0.006)+ 0.042=>0.036$

Therefore

Confidence interval is

$-0.048$

Conclusion

Given the confidence interval has zero

Therefore do not reject $H_0:P_1-P_2=0$

6 0

2 years ago

Find the first three iterates of the function f(z) = z2 + c with a value of c = 2 - 3i and an initial value of z0 = 1 + 2i.

Artist 52 [7]

Answer:

Step-by-step explanation:

We have: (I rewrote the function)

$f(z_n)={z_{n-1}} ^2+c$

Given that:

$\displaystyle c=2-3i \text{ and } z_0 = 1 + 2 i$

The first iterate will be:

$\displaystyle \begin{aligned} f(z_1)&=(z_0)^2+c \\ &=(1+2i)^2+(2-3i) \\ &= (1+4i+4i^2)+(2-3i) \\ &=1+4i-4+2-3i \\ &=-1+i \end{aligned}$

The second iterate will be:

$\begin{aligned}f(z_2) &=(z_1)^2+c\\ &=(-1+i)^2+(2-3i) \\&= (1-2i+i^2)+(2-3i) \\&=1-2i-1+2-3i \\&=2-5i \end{aligned}$

And the third iterate will be:

$\begin{aligned} f(z_3)&=(z_2)^2+c\\ &=(2-5i)^2+(2-3i) \\ &=(4-20i+25i^2)+(2-3i) \\ &=4-20i-25+2-3i \\ &=-19-23i \end{aligned}$

Hence, our answer is C.

3 0

3 years ago