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2 years ago

How many yards are in 45 inches

Mathematics

1 answer:

alexandr402 [8]2 years ago

3 0

Answer:

1.25

Step-by-step explanation:

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notsponge [240]

Streamers: 24 dollars Package of ballons: 6 dollars

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2 years ago

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What would the unit rates be for a person who jogged 500 meters in 4 minutes?

kozerog [31]

500 meters/4 minutes = 125 meters/minute

<span>4 minutes / 500 meters = 0.008 minutes/meter</span>

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2 years ago

What is the variance?

kotegsom [21]

Answer:

Variance is the squared deviation and deviation is the difference of observed value from other values

Step-by-step explanation:

Variance is the squared deviation and deviation is the difference of observed value from other values

$variance=\text({standard deviation})^2$

$Variance=\sum\frac{(x-\mu)^2}{N}$

Where, $\mu=\text{mean},N=\text{Number of sample space}$

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2 years ago

Y=x+4, 2x+y=13 solve the following system of equation. Please help

alexandr1967 [171]

Well you have to subtract -2 to both sides and it well look like Y= -2+13 and then y=x+4

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2 years ago

Suppose you are investigating the relationship between two variables, traffic flow and expected lead content, where traffic flow

prisoha [69]

Answer:

The answers is " Option B".

Step-by-step explanation:

$CI=\hat{Y}\pm t_{Critical}\times S_{e}$

Where,

$\hat{Y}=$ predicted value of lead content when traffic flow is 15.

$\to df=n-1=8-1=7$

$95\% \ CI\ is\ (463.5, 596.3) \\\\\hat{Y}=\frac{(463.5+596.3)}{2}\\\\$

$=\frac{1059.8}{2}\\\\=529.9$

Calculating thet-critical value $t_{ \{\frac{\alpha}{2},\ df \}}=-2.365$

The lower predicted value $=529.9-2.365(Se)$

$463.5=529.9-2.365(Se)\\\\529.9-463.5=2.365(Se)\\\\66.4=2.365(Se)\\\\Se=\frac{66.4}{2.365} \\\\Se=28.076$

When $99\%$ of CI use as the expected lead content: $\to 529.9\pm t_{0.005,7}\times 28.076 \\\\=(529.9 \pm 3.499 \times 28.076)\\\\=(529.9 \pm 98.238)\\\\=(529.9-98.238, 529.9+98.238)\\\\=(431.662, 628.138)\\\\=(431.6, 628.1)$

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2 years ago