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3 years ago

Which digit is in the tenths place? 48.92

2 answers:

7 0

Inside the number 48.92:
4 is in the tens place
8 is in the ones place
9 is in the tenths place
2 is in the hundredths place

Therefore, your answer is 9

hope this helps

Rama09 [41]3 years ago

3 0

9 is in the tenths place.

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4 years ago

Which will result in a perfect square trinomial (3x-5) (-3x-5)?

makvit [3.9K]

Answer:

-9x² + 25

Step-by-step explanation:

To solve, use FOIL. FOIL:

First

Outside

Inside

Last

Solve "First":

(3x)(-3x) = -9x²

Solve "Outside":

(3x)(-5) = -15x

Solve "Inside":

(-5)(-3x) = +15x

Solve "Last":

(-5)(-5) = +25

Combine like terms:

-9x² - 15x + 15x + 25

-9x² (-15x + 15x) + 25

-9x² + 25

-9x² + 25 is your answer.

8 0

3 years ago

PLS HELP :(((

Paul [167]

Hey,
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7 0

3 years ago

Which statements must be true regarding the diagram?<br> (please check more than one answer)

nikdorinn [45]

Try this option:
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4 0

3 years ago

Read 2 more answers

In a population of similar households, suppose the weekly supermarket expense for a typical household is normally distributed wi

Rina8888 [55]

Answer:

P(Y ≥ 15) = 0.763

Step-by-step explanation:

Given that:

Mean =135

standard deviation = 12

sample size n = 50

sample mean $\overline x$ = 140

Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses

Then,

$X \sim N ( \mu \sigma)$

The probability that X is greater than 140 is :

P(X>140) = 1 - P(X ≤ 140)

$P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{140-135}{12})$

$P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{5}{12})$

$P(X>140) = 1 - P( Z\leq0.42)$

From z tables,

$P(X>140) = 1 - 0.6628$

$P(X>140) = 0.3372$

Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140

Then;

$Y \sim Binomial (np)$

$Y \sim Binomial (50,0.3372)$

∴

P(Y ≥ 15) = 1- P(Y< 15)

P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )

$P(Y \geq 15) = 1 - \begin {pmatrix} ^{50}_0 \end {pmatrix} (0.3372)^0 (1-0,3372)^{50} + \begin {pmatrix} ^{50}_1 \end {pmatrix} (0.3372)^1 (1-0,3372)^{49} + \begin {pmatrix} ^{50}_2 \end {pmatrix} (0.3372)^2 (1-0,3372)^{48} +... + \begin {pmatrix} ^{50}_{50{ \end {pmatrix} (0.3372)^{50} (1-0,3372)^{0}$

P(Y ≥ 15) = 0.763

7 0

3 years ago