Find the coordinates of the points are which divides line segment!

Linda has d dollars in an account that pays 1.4% interest, compounded weekly. She withdraws w dollars. Express her first week’s

Fittoniya [83]

Answer:

If the 1.4% rate of interest is the weekly rate of interest, Then, total first week interest = 1.4(d - w)/100 = 0.014(d - w)

But if the 1.4% rate of interest is a yearly rate of interest compounded weekly, then her total first week interest = 7(d - w)/2600 = 0.00269 (d - w)

Step-by-step explanation:

The interest, I = PRT

P = initial amount of dollars in account = (deposit - withdrawal) = (d - w)

R = rate of interest = 1.4% = 0.014/week

T = time = 1 week

If the 1.4% rate of interest is the weekly rate of interest

I = PRT = (d - w) × 0.014 × 1 = 0.014(d - w)

But if the 1.4% rate of interest is a yearly rate if interest compounded weekly,

P = initial amount of dollars in account = (deposit - withdrawal) = (d - w)

R = rate of interest = 1.4% = 0.014/year

T = time = 1 week = (1/52) years

I = PRT = (d - w) × 0.014 × (1/52) = 7(d - w)/2600 = 0.00269 (d - w)

Hope this Helps!!!

7 0

4 years ago

Read 2 more answers

Audrey is buying a new car for 32,998.00. She plans to make a down payment of 4,200.00. If she makes a payment of 525 for the ne

Klio2033 [76]

Your answer would be 3.7% i believe.

4 0

3 years ago

The Fish and Game Department stocked a lake with fish in the following proportions: 30% catfish, 15% bass, 40% bluegill, and 15%

vagabundo [1.1K]

Answer:

1) $\chi^2 =\frac{(112-150)^2}{150}+\frac{(95-75)^2}{75}+\frac{(210-200)^2}{200}+\frac{(83-75)^2}{75}=16.313$

2) $p_v =P(\chi^2_{3}>16.313)=0.000978$

And we got the same decision reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts

The Chi-Square test of independence is used "to determine if there is a significant relationship between two nominal (categorical) variables". And is defined with the following statistic:

$\chi^2 =\sum_{i=1}^n \frac{(O-E)^2}{E}$

Where O rpresent the observed values and E the expected values.

State the null and alternative hypothesis

Null hypothesis: The distribution is 30% catfish, 15% bass, 40% bluegill, and 15% pike

Alternative hypothesis: The distribution is NOT 30% catfish, 15% bass, 40% bluegill, and 15% pike

The observed values are given by the table given:

Catfish =112, BAss = 95, Bluegill=210, Pike=83

Calculate the expected values

In order to calculate the expected values we can use the following formula for each cell of the table

$E = \% Grand total$

$E_{Catfish}=500*0.3=150$

$E_{Bass}=500*0.15=75$

$E_{Bluegill}=500*0.4=200$

$E_{Pike}=500*0.15=75$

Part 1: Calculate the statistic

$\chi^2 =\frac{(112-150)^2}{150}+\frac{(95-75)^2}{75}+\frac{(210-200)^2}{200}+\frac{(83-75)^2}{75}=16.313$

$\chi^2 =16.313$

Calculate the critical value

First we need to calculate the degrees of freedom given by:

$df= (categories-1)=(4-1)= 3$

Since the confidence provided is 95% the significance would be $\alpha=1-0.95=0.05$ and we can find the critical value with the following excel code: "=CHISQ.INV(0.95,3)", and our critical value would be $\chi^2_{crit}=7.815$

We can calculate also the p value:

$p_v =P(\chi^2_{3}>16.313)=0.000978$

And we got the same decision reject the null hypothesis at 5% of significance.

5 0

3 years ago

Some emoji math, there are more to be solve as well. 15 points for each. Must explain how to get the answers you got.

Maslowich

Answer:

Ladybug=7

clock= 5

bulb= 2

Step-by-step explanation:

there are a whole row of ladybugs that equal 21. 21/3=7.

on the third row, the value we need is 12, 6 won't work because we also have a bulb. so pug in 5, 5+5=10, that means the bulb is 2.

So now that we know the values, lets plug them into the first problem to see if it gives us 14... it does

So we have the correct answers

Also, you can plug in the values for the the bottom numbers and you will get those values

7 0

3 years ago

Read 2 more answers

A bit is a 0 or a 1. a bit string is a sequence of digits, all of which are either 0 or 1. for example, there are 8 possible bit

Kipish [7]

For a string length of 1 there are 2 possible strings (2^1)
for a string length of 2 there are 4 possible strings (2^2)
for a string length of 3 there are 8 possible strings (2^3)
So, for string length of 7 there are 128 possible strings (2^7)

3 0

3 years ago