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3 years ago

7

Last week Tiffany practice her facile a total of 7 hours this was 2 hours more than to practice the previous week how many hours

did Tiffany park there's the previous week

1 answer:

aleksley [76]3 years ago

8 0

Answer:5 hours

Step-by-step explanation:

7 - 2 = 5

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Deshaun deposited $4000 into an account with 5.1% interest, compounded semiannually. Assuming that no withdrawals are made, how

kompoz [17]

Answer:

A = $4652.37

Step-by-step explanation:

Given:

Initial amount (P) = $4000

Interest rate (r) = 5.1 %

n = 2 (Compounding per year)

t = 3 years

Using Compound interest formula:

$A =P(1+ r/n )^(nt)\\\\A = 4000( 1 + (0.051/2))^(2*3)\\A = $4652.37$

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3 years ago

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HELP ME PLEASEEEEEEEEE

DIA [1.3K]

Answer:

x=24

Step-by-step explanation:

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3 years ago

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Consider the function FX equals -2/3 X -24 which conclusions can be drawn about F negative 1X select two options a F negative 1X

omeli [17]

The conclusions that can be drawn about f–1(x) are:

$f^-^1(x)$ has an x-intercept of (-36, 0)
$f^-^1(x)$ has a range of all real numbers.

<h3>Calculations and Parameters:</h3>

Given:

$f(x) = \frac{-2}{3}x - 24$

$f(x) + 24 = \frac{-2}{3}x$

To find x, we would multiply both sides by -2/3

= $x= \frac{-3}{2} (f(x) + 24)$

$f^-^1(x) = -\frac{3}{2} (0) - 36$

$f^-^1(x) = -36$

Therefore, $f^-^1(x)$ has an x-intercept of (-36, 0)

Furthermore, $f^-^1(x)$ has a range of all real numbers.

Read more about real numbers here:

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5 0

2 years ago

PLZ HELP ASPPP PLZ I will mark your answer as brainlist

Natasha2012 [34]

Answer:

a(-2,4) b(-4,2) c(-8,6) d(-6,8) these should be the answers

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3 years ago

Determine whether the normal sampling distribution can be used. The claim is p < 0.015 and the sample size is n

Svetradugi [14.3K]

Complete Question

Determine whether the normal sampling distribution can be used. The claim is p < 0.015 and the sample size is n=150

Answer:

Normal sampling distribution can not be used

Step-by-step explanation:

From the question we are told that

The null hypothesis is $H_o : p = 0.015$

The alternative hypothesis is $H_a : p < 0.015$

The sample size is n= 150

Generally in order to use normal sampling distribution

The value $np \ge 5$

So

$np = 0.015 * 150$

$np = 2.25$

Given that $np < 5$ normal sampling distribution can not be used

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3 years ago