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artcher [175]
3 years ago
9

Express the location of the point on the number line as both a fraction and a decimal.

Mathematics
1 answer:
insens350 [35]3 years ago
3 0
Decimal: 0.3
Fraction: 3/10
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164 is 55% of what number?<br><br> Please somebody help me
monitta
This is how you solve it; but first of all convert the percent into a decimal;55%= 0.55

0.55x = 164 (divide by 0.55 on both sides)
x = 164/0.55 
x = 298.18 (rounded to the nearest hundredth)
so 164 is 55% of 298.18(rounded)
8 0
3 years ago
Help please help me please help please please help me please
GarryVolchara [31]
The answer is no. the first equation is not right
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‼️ PLEASEEE HELPPPP ‼️
V125BC [204]

Answer: answer is in the photo

Step-by-step explanation:

8 0
3 years ago
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Find the length of CA to the nearest tenth of a foot
hoa [83]

Answer:

CA ≈ 3.1 ft

Step-by-step explanation:

Using the tangent ratio in the right triangle

tan40° = \frac{opposite}{adjacent} = \frac{BC}{CA} = \frac{2.6}{CA} ( multiply both sides by CA )

CA × tan40° = 2.6 ( divide both sides by tan40° )

CA = \frac{2.6}{tan40} ≈ 3.1 ft ( to the nearest tenth )

3 0
3 years ago
According to a study done by Wakefield Research, the proportion of Americans who can order a meal in a foreign language is 0.47.
UNO [17]

Answer:

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

Step-by-step explanation:

We are given that according to a study done by Wake field Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

Suppose a random sample of 200 Americans is asked to disclose whether they can order a meal in a foreign language.

<em>Let </em>\hat p<em> = sample proportion of Americans who can order a meal in a foreign language</em>

The z-score probability distribution for sample proportion is given by;

          Z = \frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion

p = population proportion of Americans who can order a meal in a foreign language = 0.47

n = sample of Americans = 200

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is given by = P( \hat p > 0.50)

  P( \hat p > 0.06) = P( \frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } > \frac{0.5-0.47}{\sqrt{\frac{0.5(1-0.5)}{200} } } ) = P(Z > 0.85) = 1 - P(Z \leq 0.85)

                                                               = 1 - 0.80234 = <u>0.19766</u>

<em>Now, in the z table the P(Z  </em>\leq <em>x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.85 in the z table which has an area of 0.80234.</em>

Therefore, probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

4 0
4 years ago
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