What is 1,357 times 2

Orlov [11]

Answer: Can you tell me how do you want it answered?

Step-by-step explanation:

5 0

2 years ago

1) A librarian noticed that 60% of seventh graders checked out fantasy books. About how many of 240 seventh graders would checko

katovenus [111]

We have been given that a librarian noticed that 60% of seventh graders checked out fantasy books. We are asked to find the number of seventh graders from 240 seventh graders, who would checkout fantasy books.

To solve our given problem, we need to find 60% of 240.

$\text{Seventh graders, who would checkout fantasy books}=240\times \frac{60}{100}$

$\text{Seventh graders, who would checkout fantasy books}=240\times 0.60$

$\text{Seventh graders, who would checkout fantasy books}=144$

Therefore, 144 seventh graders would checkout fantasy books.

3 0

3 years ago

Read 2 more answers

1.

LUCKY_DIMON [66]

Answer:

a 72

b 7

Part C)

___7_|_35+63_

_____|__9+5__

35+63=7(9+5)=714=98 or 35+63=98

Step-by-step explanation:

5 0

3 years ago

16. Construct 3 equations starting with x = 5.

strojnjashka [21]

TO FIND:  construct 3 equation starting with x=5?

SOLUTION: 

The equation is in the form of variable and constant equating.

Let the equation be x=5.

TO CREATE AN EQUATION,

FIRST WE ADD 5 BOTH SIDe,

x + 5=5+5

=x + 5=10

SECOND WE SUBRACT 5 BOTH SIDE,

x — 5=5—5

=x—5=0

THIRD WE MULTIPLY 5 BOTH SIDE,

5x=5×5

=5x=25

well u got the answer!

3 0

3 years ago

A study was recently conducted to estimate the mean cholesterol for adult males over the age of 55 years. From a random sample o

Komok [63]

Answer: Correct critical value = 2.2622

Step-by-step explanation:

Confidence interval for population mean when population standard deviation is unknown:

$\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$ , where $\overline{x}$ = sample mean, n= sample size, s= sample standard deviation, $t_{\alpha/2}$ = two-tailed t value.

As per given: n= 10

degree of freedom : df = n-1=9

$\overline{x}=242.6\\\\s=73.33\\\\\alpha=5\%=0.05$

Critical t-value : $t_{\alpha/2,df}=t_{0.05/2,9}=t_{0.025,9}=2.2622$

So, the 95 percent confidence interval estimate for the mean :

$242.6\pm (2.2622)\dfrac{73.33}{\sqrt{10}}\\\\=242.6\pm (2.2622)(23.19)\\\\\approx242.6\pm52.46\\\\=(242.6-52.46,\ 242.6+52.46)\\\\=(190.14,\ 295.06)$

The 95 percent confidence interval estimate for the mean:(190.14, 295.06)

8 0

2 years ago