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2 years ago

Round 4.35 to the nearest tenths

Mathematics

2 answers:

bonufazy [111]2 years ago

5 0

It will be 4.4 when rounded up.

JulijaS [17]2 years ago

3 0

4.35 rounded to the nearest tenths would be 4.4.

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If you pay $35 for a drug and the selling price is $50 what is the mark-up of this drug

Daniel [21]

Answer:

15$

Step-by-step explanation:

15$

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2 years ago

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Solve the linear equation:<br><br> x-y=7<br><br> 0.5x+y=5

evablogger [386]

Add the equations.

x-y%2B0.5x%2By=7%2B5

1.5x=12

highlight%28x=8%29

Now use either equation to solve for y.

8-y=7

highlight%28y=1%29

x=8

y=1

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2 years ago

Which are equivalent to 3^2 x 3^4? check all that apply.

Andreas93 [3]

equivalent

1) 3^6

4) 3^-4 · 3^10

8) (3 · 3) · (3 · 3 · 3 · 3)

Step-by-step explanation:

$= {3}^{2} \times {3}^{4}$

$= {3}^{(2 + 4)}$

$= {3}^{6}$

equivalent

1) 3^6

4) 3^-4 · 3^10

8) (3 · 3) · (3 · 3 · 3 · 3)

8 0

2 years ago

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In a manufacturing process a random sample of 9 bolts manufactured has a mean length of 3 inches with a standard deviation of .3

Serggg [28]

Answer:

The correct option is (C) (2.769, 3.231).

Step-by-step explanation:

The confidence interval for mean when the standard deviation is not known is:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\frac{s}{\sqrt{n}}$

Given:

$\bar x = 3\\s=0.3\\n=9\\\alpha =1-0.95=0.05$

Compute the critical value as follows:

$t_{\alpha/2, (n-1)}=t_{0.05/2, (9-1)}=t_{0.025, 8}=2.31$

**Use a <em>t</em>-table.

The 95% confidence interval for true mean length of the bolt is:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\frac{s}{\sqrt{n}}\\=3\pm 2.31\times \frac{0.30}{\sqrt{9}}\\ =3\pm 0.231\\=(2.769, 3.231)$

Thus, the 95% confidence interval for true mean length of the bolt is (2.769, 3.231).

The correct option is (C).

7 0

2 years ago

Sean used the $1,200 he got from his graduation party to open a savings account. If the account earns 1% interest each month and

Greeley [361]

Answer:

$\$1,920$

Step-by-step explanation:

we know that

The simple interest formula is equal to

$A=P(1+rt)$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest

t is Number of Time Periods

in this problem we have

$t=5*12=60\ months\\ P=\$1,200\\r=0.01$

substitute in the formula above

$A=\$1,200(1+0.01*60)$

$A=\$1,200(1.6)=\$1,920$

6 0

2 years ago