MATHEMATICS COLLEGE

Estimate the product 4x279

Answers

Answer 1

Answer:

Answer:

116

Step-by-step explanation:

es 116 porque lo cheque

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Can someone help me solve this please?

The two triangles shown are similar. Find the value of
a
b
2.1
1.4

Answers

Answer:a 2.1 similar to 1.4 is 5.6 n answer to your question

Step-by-step explanation:

Answer:

1.5

Step-by-step explanation:

In tilapia, an important freshwater food fish from Africa, the males actively court females. They have more incentive to court a female who has already laid all of her eggs, but can they tell the difference? an experiment was done to measure the male tilapia's response to the smell of female fish. Water containing feces from females that were either pre-ovulatory (they still had eggs) or post-ovulatory (they had already laid their eggs) was washed over the gills of males hooked up to an electro-olfactogram machine which measured when the senses of the males were excited. The amplitude of the electro-olfactogram was used as a measure of the excitability of the males in the two different circumstances. Six males were exposed to the scent of pre-ovulatory females; their readings average 1.51 with a standard deviation of .25. Six different males were exposed to post-ovulatory females; their average readings of 0.87 with standard deviation is .31. Assume that the electro-olfactogram readings were approximately normally distributed within the groups.(A) test for a difference in the excitability of the males with exposure to these two types of females
(B) what is the estimated average difference in electro-olfactogram readings between the two groups? What is the 95% confidnece limit for the difference between population means?

Answers

Answer:

a) $t=\frac{1.51-0.87}{\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}}=3.936$

"=T.INV(1-0.025,10)", and we got $t_(critical)=\pm 2.28$

Statistical decision

Since our calculated value is higher than our critical value, $z_(calc)=3.936>2.28=t_(critical)$ , we have enough evidence to reject the null hypothesis at 5% of significance.

b) $(\bar X_1 -\bar X_2) \pm t_(\alpha/2)\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}$

The degrees of freedom are given:

$df = n_1 + n_2 -2 = 6+6-2 = 10$

$(1.51 -0.87) - 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 0.269$

$(1.51 -0.87) + 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 1.010$

Step-by-step explanation:

Part a

Data given and notation

$\bar X_(1)=1.51$ represent the mean for scent of pre ovulatory

$\bar X_(2)=0.87$ represent the mean for post ovolatory

$s_(1)=0.25$ represent the sample standard deviation for preovulatory

$s_(2)=0.31$ represent the sample standard deviation for postovulatory

$n_(1)=6$ sample size for the group preovulatory

$n_(2)=6$ sample size for the group postovulatory

z would represent the statistic (variable of interest)

$p_v$ represent the p value

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean's are different, the system of hypothesis would be:

H0: $\mu_(1) = \mu_(2)$

H1: $\mu_(1) \neq \mu_(2)$

If we analyze the size for the samples both are lower than 30, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_(1)-\bar X_(2)}{\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We have all in order to replace in formula (1) like this:

$t=\frac{1.51-0.87}{\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}}=3.936$

Find the critical value

We find the degrees of freedom:

$df = n_1 + n_2 -2 = 6+6-2 = 10$

In order to find the critical value we need to take in count that we are conducting a two tailed test, so we are looking for thwo values on the t distribution with df =10 that accumulates 0.025 of the area on each tail. We can us excel or a table to find it, for example the code in Excel is:

"=T.INV(1-0.025,10)", and we got $t_(critical)=\pm 2.28$

Statistical decision

Since our calculated value is higher than our critical value, $z_(calc)=3.936>2.28=t_(critical)$ , we have enough evidence to reject the null hypothesis at 5% of significance.

Part b

For this case the confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm t_(\alpha/2)\sqrt{(s^2_(1))/(n_(1))+(s^2_(2))/(n_(2))}$

The degrees of freedom are given:

$df = n_1 + n_2 -2 = 6+6-2 = 10$

$(1.51 -0.87) - 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 0.269$

$(1.51 -0.87) + 2.28\sqrt{(0.25^2)/(6)+(0.31^2)/(6)}= 1.010$

Solve the right triangle ABC, with C = 90.00◦ , a = 15.21 cm, b = 17.34 cm. Round to two decimal places.

Answers

Answer:

the hypotenuse side, c = 23.1 cm

angle A = 41.26 ⁰

angle B = 48.74 ⁰

Step-by-step explanation:

Given;

first leg of the right triangle, a = 15.21 cm

second leg of the right triangle, b = 17.34 cm

Angle C = 90 ⁰

The missing parameters;

the hypotenuse side = c

angle A

angle B

Use Pythagoras theorem to calculate the missing side "c", which is the hypotenuse

c² = a² + b²

c² = (15.21)² + (17.34)²

c² = 532.02

c = √532.02

c = 23.1 cm

The missing angle A is calculated as;

$tan(A) = (a)/(b) \n\ntan(A) = (15.21)/(17.34) \n\ntan(A) = 0.8772 \n\nA = tan^(-1) (0.8772)\n\nA = 41.26^0$

The missing angle is calculated as;

B = 90⁰ - A

B = 90⁰ - 41.26⁰

B = 48.74⁰

You have a 1 kilogram bag of sugar You use 435 grams to fill the sugar bowls
How much sugar is left in the bag

Answers

The answer is 565g. Unless if there are more bowls, you just minus 435 from 565.

There is 1,000 grams in a kilogram. I just took away 435 from 1,000.
1,000 - 435 = 565 grams left in the bag.

Y-3=2(x+1) please please help

Answers

Answer:

y=2x+5

Step-by-step explanation:

distribute 2 to x and 1, do this by multiplying each.

you should get y-3=2x+2.

then you have to move the -3 from the left to the right. but when you do this you have to change the -3 to a positive 3. then combine 3 and 2.

you then get the answer, y=2x+5.

Joe and Susan are both 40 years old and hope to have enough money saved to retire by the time they're 65. They deposit $6,000 eachyear into an account that pays 4% interest compounded annually. Use this information to complete the table.
Formula to Use
Account Balance in 20 years
Total Interest Earned
present value of an annuity
future value of a lump sum
future value of an annuity
$99,875
$112,253
$149,875
$249,875
$286,363

Answers

Account balance In 20 years is $249,875 and total interest earned is $99,875.

What is the future value?

The FV formula assumes a constant rate of growth and a single up-front payment left untouched for the duration of the investment. The FV calculation can be done one of two ways, depending on the type of interest being earned.

We have,

P = $6,000

r = 0.04

n = 1

t = 25

We know the formula for future value

FV= PV $(1 + r)^(n)$ .

where:

P=Investment amount

R=Interest rate

T=Number of years

According to the question

FV=$6,000×(1+0.04

FV = $249,875

To find the total interest earned, first find the total value of the deposit payments:$6,000 annual payments for 25 years = 6,000 × 25= $150,000

Total interest earned = FV - annually payment for 25 year = $249,875 - $150,000 = $99,875

To learn more about future value from here

brainly.com/question/14860893

#SPJ2

Answer:

Formula: Future value of annuity

In 20 years: $249,875

Total Interest: $99,875

Step-by-step explanation:

Use the formula for future value:

FV =

P = $6,000

r = 0.04

n = 1

t = 25

FV =

FV = $249,875

To find the total interest earned, first find the total value of the deposit payments:

$6,000 annual payments for 25 years = 6,000 • 25

= $150,000

(I copied that explanation from somewhere else so its a little messed up. I just couldn't explain it properly. Hope this helps!)

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