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4 years ago

2007 divided by j = 223

Mathematics

1 answer:

Effectus [21]4 years ago

8 0

J = 9 because 2007 / 223 = 9, and when you multiply 223 x 9, you get 2007

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Which answer choice shows 21.97 written in expanded form? A. 2 + 1 + 9 + 7 B. 20 + 1 + 0.9 + 0.7 C. 20 + 10 + 0.9 + 0.07 D. 20 +

Simora [160]

Answer:

20 + 1 + 0.9 + 0.07

Step-by-step explanation:

Given the number 21.97 ;

To wite in expanded form:

Tens ___ Unit ____ tenth ____ hundredth

2_______ 1 _______ 9_________ 7

2 = 20

1 = 1

9 = 0.9

7 = 0.07

Hence,

20 + 1 + 0.9 + 0.07

6 0

4 years ago

A triangle has an area of 72 square inches. If the base of the triangle has a length of 18 inches, what is the height of the tri

nikdorinn [45]

Answer:

height is 4 in

Step-by-step explanation:

total area 72

divide area by base length

72/18= 4

72 total area

18 base length

4 height

6 0

4 years ago

Find the value of h in this triangle. Please and Thank you!

Dahasolnce [82]

Answer:

h= 10m

Step-by-step explanation:

1/2x12x18.2=109.2m squared

then u divide it by 21.8

109.2m2/21.8m/ 1/2=10m

h=10m

8 0

3 years ago

A simple random sample from a population with a normal distribution of 103 body temperatures has x overbarequals98.90degrees Upp

Shtirlitz [24]

Answer:

$98.90-1.984\frac{0.62}{\sqrt{103}}=96.98$

$98.90+ 1.984\frac{0.62}{\sqrt{103}}=100.82$

The 95% confidence interval would be given by (96.98;100.82)

Step-by-step explanation:

Information given

$\bar X=98.90$ represent the sample mean

$\mu$ population mean (variable of interest)

s=0.62 represent the sample standard deviation

n=103 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom are given by:

$df=n-1=103-1=102$

Since the Confidence is 0.95 or 95%, the significance is $\alpha=0.05$ and $\alpha/2 =0.025$ , and the critical value for this case would be $t_{\alpha/2}=1.984$

And replacing we got:

$98.90-1.984\frac{0.62}{\sqrt{103}}=96.98$

$98.90+ 1.984\frac{0.62}{\sqrt{103}}=100.82$

The 95% confidence interval would be given by (96.98;100.82)

8 0

3 years ago

A certain large manufacturing facility produces 20,000 parts each week. The manager of the facility estimates that about 1% of t

Slav-nsk [51]

Answer:

198 is the variance for the number of defective parts made each week.

Step-by-step explanation:

We are given the following in the question:

Number of parts produced each week = 20,000

Percentage of defective parts = 1%

We have to calculate the variance for the number of defective parts made each week.

We treat defective part as a success.

P(Defective part) = 1% = 0.01

$p = 0.01$

Then the number defective parts follows a binomial distribution .

Formula for variance =

$\sigma^2 = np(1-p) = 20000(0.01)(1-0.01) = 198$

Thus, 198 is the variance for the number of defective parts made each week.

8 0

4 years ago